{"version":1,"lectureId":"01M14TZWMWWK33R8DXJ7Z859ME","attempt":1,"publication":{"slug":"gauss-law-and-what-flux-is-counting","title":"Electric Flux and Gauss’s Law: How Symmetry Reveals Electric Fields","subject":"physics","summary":"Electric flux begins as a signed count of field lines crossing a surface and becomes the surface integral in Gauss’s law. The lecture develops that law from a point charge, then uses spherical, cylindrical, and planar symmetry to find the fields of a uniformly charged solid sphere, an infinite line, and an infinite sheet. It closes with electrostatic conductors, surface charge, shielding, and the protection provided by a metal car body during lightning.","metaDescription":"Build electric flux and Gauss’s law visually, apply symmetry to spheres, lines and sheets, then explain conductors and lightning shielding.","transcript":"Place one positive point charge in empty space. At every surrounding point, its electric field points directly away from the charge. These blue arrows represent field lines, a drawing that lets us follow the direction of the field through space. The picture is three dimensional. Every direction away from the charge is equivalent, and no radial direction is preferred over another. Now surround the charge with a small sphere. Every blue line leaving the charge crosses that sphere once, from its inside to its outside. If we use the lines as a counting picture, the outward flux is the number of crossings. The lines are not physical threads, and their drawn number is arbitrary. Still, the crossing picture captures something real: a surface facing the field receives positive flux, and a closed surface surrounding the source intercepts the whole outward pattern. Replace the small sphere with a larger one. The same radial lines cross it, so the crossing count has not changed. They are farther apart now, which is the picture's way of saying that the electric field is weaker. The radius grew, and the sphere's area grew as radius squared. But the field of a point charge weakened as one over radius squared. Twice the radius gives four times the area and one quarter of the field strength. Their product stays fixed. A sphere is not essential. Dent the surface here, bulge it there, and keep the charge enclosed. Every ray that leaves the charge still has to cross the closed boundary. The local crossing angles and local field strengths change, but the total outward count does not. For a sphere of radius r, the point-charge field has magnitude one over four pi epsilon zero, times q over r squared. The sphere's area is four pi r squared. This is exactly the geometrical growth that compensates for the inverse-square weakening. Multiply field strength by area. The r squared in the area cancels the r squared in the denominator, and four pi cancels as well. What remains is q divided by epsilon zero. It has no radius in it. The calculation therefore agrees with the crossing picture: every surrounding sphere receives the same total outward electric flux. The lumpy surface needs a more careful sum because its field strength and angle vary from patch to patch. But once those local contributions are counted correctly, the same enclosed charge gives the same total flux. That invariant total is the idea Gauss's law will make exact. The next step is to define what one small crossing contributes, including the angle at which the field meets the surface. Zoom in on one small piece of a surface. Its area is delta A, and the yellow arrow is the outward unit normal, the direction perpendicular to that patch. Let the electric field meet the patch at an angle theta from the normal. The field crosses most effectively when it points along the normal. The patch's flux is E dot n hat times delta A. Equivalently, it is field strength times area times cosine theta. The cosine selects the normal component of the field. If the field runs parallel to the patch, theta is ninety degrees and the cosine is zero. Lines may skim along the surface, but none cross it, so that part contributes no flux. If the field points outward, the dot product is positive. If it points inward, the dot product is negative. Flux therefore counts signed crossings, outward minus inward. A curved surface is built from many small patches. Add the local dot products over all of them. As the patches become arbitrarily small, the sum becomes a surface integral. This is the precise definition of electric flux. Unlike the number of lines an artist draws, the integral has a fixed numerical meaning. Return to a point charge and a spherical surface around it. Symmetry makes the field radial, so it points along the outward normal everywhere on the sphere. Symmetry also makes the field strength the same at every point of a sphere with radius r. We can therefore take E outside the integral, leaving the total area four pi r squared. Insert the inverse-square point-charge field. The area growth cancels the field's weakening, leaving q divided by epsilon zero. Now deform the closed surface without moving it across the charge. Some patches tilt, some move closer, and others move farther away. The local dot products change, but the net signed crossing count cannot change. Several charges simply add their fields. A charge inside contributes its full outward flux. A charge outside sends as much flux into the closed surface as it sends back out, so its net contribution is zero. The result is Gauss's law: the flux through any closed surface equals the net enclosed charge divided by epsilon zero. The surface may be imaginary, irregular, or placed wherever we choose. Written in full, the closed-surface integral of E dot n hat equals enclosed charge divided by epsilon zero. Notice what the law does and does not say. It always gives total flux from enclosed charge. It gives the electric field itself only when symmetry makes the field's direction and magnitude simple on a carefully chosen surface. That choice is the real technique. For a spherical charge distribution we will choose a sphere. For a line we will choose a cylinder. For a sheet we will choose a pillbox. In every case, the source geometry chooses the useful Gaussian surface. Take a solid insulating sphere of radius R, with total charge Q spread uniformly throughout its volume. Because the distribution looks the same after any rotation about its centre, the electric field must point radially. A short turn confirms the geometry. There is no preferred direction around the sphere, so at a fixed distance from the centre every point must have the same field magnitude. First ask for the field outside the charge distribution. Choose a spherical Gaussian surface of radius r greater than R, centred on the same point. On this Gaussian sphere, the field is everywhere normal to the surface and has one constant magnitude E of r. The flux is therefore E times four pi r squared. The Gaussian surface encloses the entire charged body, so the enclosed charge is Q. Gauss's law sets the flux equal to Q over epsilon zero. Solving for E gives the familiar inverse-square field. Outside a spherically symmetric distribution, the field is exactly the same as if all the charge were concentrated at the centre. Gauss's law makes that statement exact, not merely approximate. Now move the Gaussian sphere inside the charged body. Its radius r is less than R. The field still has spherical symmetry, but the surface now encloses only part of the total charge. Uniform volume charge density means total charge divided by total volume. The density rho is Q over four thirds pi R cubed. The smaller Gaussian sphere has volume four thirds pi r cubed. Multiply that volume by rho to find the charge it encloses. The common factors cancel, leaving Q times r cubed over R cubed. This fraction is simply the fraction of the charged volume lying inside the Gaussian surface. Gauss's law again says E times four pi r squared equals enclosed charge over epsilon zero. One power of r survives after division. The interior field is proportional to r. It is zero at the centre, where all directions balance, and it grows linearly as the Gaussian sphere encloses more charge. Put the two regions on one graph. Distance is measured in units of R, and field strength in units of its value at the surface. Inside, the field rises in a straight line from zero. In normalized form, E over E at the surface equals r over R. Enclosed charge grows as r cubed, while Gaussian area grows as r squared, leaving one power of r. At r equals R, the inside and outside formulas agree. There is no jump in the field because the charge fills a volume rather than sitting in an infinitesimally thin surface layer. Outside, E over the surface field equals R squared over r squared. The enclosed charge has stopped growing, while the Gaussian area continues to grow as r squared. The method was the same in both regions: use the source symmetry to choose a concentric sphere, make E constant on that surface, and then count only the charge actually enclosed. Now stretch the charge distribution into an ideal infinite line with uniform charge per length lambda. Translation along the line changes nothing, and rotation around it changes nothing. Those symmetries force the electric field to point directly away from the line. Its magnitude can depend on perpendicular distance r, but not on position along the line or angle around it. Choose a cylinder centred on the line. Its curved side is everywhere the same distance r from the charge, so E has one constant magnitude there. Let the solid settle into view. The charged line is the cylinder's axis, and the blue field arrows point through its curved wall. On the two end caps, the outward normals point along the line, while the electric field points radially away from it. Their dot product is zero, so the caps contribute no flux. Only the curved side contributes. Its area is circumference two pi r times length L, so the flux is E times two pi r L. The cylinder encloses a length L of line charge. Charge per length lambda times L gives enclosed charge lambda L. Apply Gauss's law. The length L appears on both sides and cancels. Solving leaves lambda over two pi epsilon zero r. The line field falls as one over r, not one over r squared. As the cylinder expands, its relevant area per unit length grows only in proportion to r. Next spread charge uniformly across an ideal infinite sheet, with surface charge density sigma. Sliding anywhere within the sheet cannot change the field. Rotating the sheet within its own plane also changes nothing. The only distinguished direction is perpendicular to the sheet, so the field must point normally away on both sides. Choose a short cylindrical pillbox that straddles the sheet. Its flat caps are parallel to the charge distribution, and its curved wall joins them. A small turn shows the construction. The field passes through the two caps, while it runs parallel to the curved wall. The curved wall contributes zero flux because its normal lies within the sheet while E is perpendicular to it. Each cap contributes E A, so the total flux is two E A. The pillbox encloses sheet area A, so it encloses charge sigma A. Gauss's law gives two E A equals sigma A over epsilon zero. The cap area cancels, leaving E equals sigma over two epsilon zero. There is no distance in the answer. For an ideal infinite sheet, moving the caps farther away does not spread a fixed bundle over a growing area. The same cap area intercepts the same flux. Put the three geometries together. A point spreads flux over a sphere whose area grows as r squared, so its field has the form a constant over r squared. A line spreads flux over the curved wall of a cylinder. Per unit length, that area grows as r, so the field has the form a constant over r. A sheet sends flux through two equal caps. Their area does not change when the pillbox grows taller, so the ideal sheet field is constant. Gauss's law was identical in all three cases. What changed was the symmetry, and symmetry determined the surface on which E became constant and the unwanted pieces contributed zero. A conductor contains mobile charge. If an electric field existed inside the metal, that charge would feel a force and begin to move. A state with moving charge is not electrostatic equilibrium. The mobile charge rearranges on an extremely short time scale. Its own electric field opposes the field that drove the motion, and equilibrium is reached only when the net electric field inside the conducting material is zero. That statement comes from the physics of mobile charge, not from Gauss's law alone. Gauss's law now tells us an important consequence of the zero field. Draw any closed Gaussian surface lying completely inside the conducting material. Because E is zero at every point on it, its electric flux is zero. Gauss's law then says that the net charge enclosed by every such interior surface is zero. Excess charge therefore cannot remain distributed through the bulk of the metal in electrostatic equilibrium. It moves outward and comes to rest on the conductor's surface. Outside the conductor, those surface charges can produce an electric field. Inside the conducting material, they have arranged themselves so that their combined field cancels. The surface distribution need not be uniform. Charge crowds more strongly near sharp points, where the surrounding field can become especially large. But the equilibrium condition inside the metal remains E equals zero. Now replace the simple conductor with the metal body of a hard-top car. The body forms a conducting shell around the passenger compartment. Suppose lightning strikes the roof. The lightning delivers charge and a large current to the exterior metal. The charge spreads over the outside, and the current finds conducting paths along the exterior body. The metal shell carries the dangerous electrical disturbance around the passenger space rather than through it. The passenger compartment is therefore close to one electric potential, with a strongly reduced electric field inside. This shielding behavior is often called the Faraday-cage effect. The protection comes from the continuous metal shell, not primarily from the rubber tires. During a storm, occupants should remain inside with windows closed and avoid touching metal parts connected to the exterior. The same principle is used deliberately in shielded rooms, cable coverings, and metal enclosures around sensitive electronics. Conductors rearrange charge so that their protected interiors experience very little electric field. Three ideas organize the whole lecture. First, electric flux is the signed amount of electric field crossing a surface, and the flux through a closed surface counts net enclosed charge. Second, Gauss's law becomes a field-solving tool only when symmetry chooses a useful surface: a sphere for spherical charge, a cylinder for a line, and a pillbox for a sheet. Third, mobile charge in a conductor rearranges until the field inside the metal is zero. Excess charge lives on the surface, and a closed metal body redirects an external electrical disturbance around its interior. Flux made the counting picture precise. Gauss's law connected that count to charge. Symmetry turned the law into three electric fields, and electrostatic equilibrium turned it into protection inside a conductor.","watch":{"version":1,"scenes":[{"title":"Flux That Refuses to Change","start":0,"end":177.10831249999995,"objects":{"charge":"a Point [red] labelled \"q\" drawn in space (location=(0.0, 0.0, 0.0))","equation_heading":"a Heading that says \"Weakening Field, Growing Area\"","heading":"a Heading that says \"A Point Charge and Its Field\"","large_sphere":"a Sphere [green] drawn in space (radius=1.65, opacity=0.14)","lumpy":"a Surface [magenta] drawn in space (function=<function>, u_range=(0.0, 3.141592653589793), v_range=(0.0, 6.283185307179586))","rays":"a Vector [blue] drawn in space (start=(0.18, 0.0, 0.0), end=(2.35, 0.0, 0.0))","rays_10":"a Vector [blue] drawn in space (start=(-0.12725999999999998, -0.12725999999999998, 0.0), end=(-1.6614499999999999, -1.6614499999999999, 0.0))","rays_11":"a Vector [blue] drawn in space (start=(0.12725999999999998, 0.0, 0.12725999999999998), end=(1.6614499999999999, 0.0, 1.6614499999999999))","rays_12":"a Vector [blue] drawn in space (start=(-0.12725999999999998, 0.0, 0.12725999999999998), end=(-1.6614499999999999, 0.0, 1.6614499999999999))","rays_13":"a Vector [blue] drawn in space (start=(0.0, 0.12725999999999998, -0.12725999999999998), end=(0.0, 1.6614499999999999, -1.6614499999999999))","rays_14":"a Vector [blue] drawn in space (start=(0.0, -0.12725999999999998, -0.12725999999999998), end=(0.0, -1.6614499999999999, -1.6614499999999999))","rays_2":"a Vector [blue] drawn in space (start=(-0.18, 0.0, 0.0), end=(-2.35, 0.0, 0.0))","rays_3":"a Vector [blue] drawn in space (start=(0.0, 0.18, 0.0), end=(0.0, 2.35, 0.0))","rays_4":"a Vector [blue] drawn in space (start=(0.0, -0.18, 0.0), end=(0.0, -2.35, 0.0))","rays_5":"a Vector [blue] drawn in space (start=(0.0, 0.0, 0.18), end=(0.0, 0.0, 2.35))","rays_6":"a Vector [blue] drawn in space (start=(0.0, 0.0, -0.18), end=(0.0, 0.0, -2.35))","rays_7":"a Vector [blue] drawn in space (start=(0.12725999999999998, 0.12725999999999998, 0.0), end=(1.6614499999999999, 1.6614499999999999, 0.0))","rays_8":"a Vector [blue] drawn in space (start=(-0.12725999999999998, 0.12725999999999998, 0.0), end=(-1.6614499999999999, 1.6614499999999999, 0.0))","rays_9":"a Vector [blue] drawn in space (start=(0.12725999999999998, -0.12725999999999998, 0.0), end=(1.6614499999999999, -1.6614499999999999, 0.0))","result":"a Math [text] that says \"$Phi_E = frac(q, epsilon_0)$\"","small_sphere":"a Sphere [yellow] drawn in space (radius=0.9, opacity=0.18)","space":"an Axes3D (x_range=(-2.6, 2.6), y_range=(-2.6, 2.6), z_range=(-2.6, 2.6))","work":"a Derivation [text] that says \"$E(r) = frac(1, 4 pi epsilon_0) frac(q, r^2) \\ A(r) = 4 pi r^2 \\ E(r) A(r) = frac(1, 4 pi epsilon_0) frac(q, r^2) 4 pi r^2 \\ Phi_E = frac(q, epsilon_0)$\""},"beats":[{"start":0,"say":"Place one positive point charge in empty space. At every surrounding point, its electric field points directly away from the charge. These blue arrows represent field lines, a drawing that lets us follow the direction of the field through space.","live":[],"does":[[0,"heading is shown on the screen, written out."],[0,"space is shown on the screen, written out."],[0,"charge is shown on the screen, written out."],[0,"rays is shown on the screen, written out."],[0,"rays_2 is shown on the screen, written out."],[0,"rays_3 is shown on the screen, written out."],[0,"rays_4 is shown on the screen, written out."],[0,"rays_5 is shown on the screen, written out."],[0,"rays_6 is shown on the screen, written out."],[0,"rays_7 is shown on the screen, written out."],[0,"rays_8 is shown on the screen, written out."],[0,"rays_9 is shown on the screen, written out."],[0,"rays_10 is shown on the screen, written out."],[0,"rays_11 is shown on the screen, written out."],[0,"rays_12 is shown on the screen, written out."],[0,"rays_13 is shown on the screen, written out."],[0,"rays_14 is shown on the screen, written out."]]},{"start":15.5075,"say":"The picture is three dimensional. 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Twice the radius gives four times the area and one quarter of the field strength. Their product stays fixed.","live":["space","heading","charge","rays","rays_2","rays_3","rays_4","rays_5","rays_6","rays_7","rays_8","rays_9","rays_10","rays_11","rays_12","rays_13","rays_14","large_sphere"],"does":[[85.768,"large_sphere is indicated — a transient flash."]]},{"start":88.03999999999999,"say":"A sphere is not essential. Dent the surface here, bulge it there, and keep the charge enclosed. Every ray that leaves the charge still has to cross the closed boundary. 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Lines may skim along the surface, but none cross it, so that part contributes no flux.","live":["local_formula","patch_picture","local_heading","patch","normal_arrow","field_arrow","angle"],"does":[[211.23031249999997,"parallel_field is shown on the screen, written out."],[216.75631249999995,"parallel_field is indicated — a transient flash."],[221.49331249999994,"parallel_field is hidden from the screen."]]},{"start":222.09331249999997,"say":"If the field points outward, the dot product is positive. If it points inward, the dot product is negative. Flux therefore counts signed crossings, outward minus inward.","live":null,"does":[[223.38231249999995,"normal_arrow is indicated — a transient flash."],[234.37681249999997,"patch_picture moves to a new place on the board."],[234.37681249999997,"local_formula is hidden from the screen — left the board."],[234.37681249999997,"local_heading is hidden from the screen — left the board."]]},{"start":235.57681249999996,"say":"A curved surface is built from many small patches. Add the local dot products over all of them.","live":["patch_picture","patch","normal_arrow","field_arrow","angle"],"does":[[235.57681249999996,"sum_heading is shown on the screen, written out."],[239.31531249999995,"sum_work is shown on the screen, written out."]]},{"start":242.80631249999993,"say":"As the patches become arbitrarily small, the sum becomes a surface integral. This is the precise definition of electric flux. Unlike the number of lines an artist draws, the integral has a fixed numerical meaning.","live":["patch_picture","patch","normal_arrow","field_arrow","angle","sum_heading"],"does":[[247.33431249999995,"sum_work is shown on the screen, written out."],[257.50431249999997,"patch_picture is hidden from the screen — left the board."],[257.50431249999997,"patch is hidden from the screen — patch_picture left the board."],[257.50431249999997,"normal_arrow is hidden from the screen — patch_picture left the board."],[257.50431249999997,"field_arrow is hidden from the screen — patch_picture left the board."],[257.50431249999997,"angle is hidden from the screen — patch_picture left the board."],[257.50431249999997,"sum_heading is hidden from the screen — left the board."],[257.50431249999997,"sum_work is hidden from the screen — left the board."]]},{"start":258.10431249999993,"say":"Return to a point charge and a spherical surface around it. Symmetry makes the field radial, so it points along the outward normal everywhere on the sphere.","live":[],"does":[[258.10431249999993,"sphere_axes is shown on the screen, written out."],[258.10431249999993,"source is shown on the screen, written out."],[260.6123124999999,"sphere_axes moves to a new place on the board."],[260.6123124999999,"point_work is shown on the screen, written out."],[266.92831249999995,"shell is shown on the screen, faded in."]]},{"start":268.44481249999995,"say":"Symmetry also makes the field strength the same at every point of a sphere with radius r. We can therefore take E outside the integral, leaving the total area four pi r squared.","live":["sphere_axes","source","shell"],"does":[[278.1623125,"point_work is shown on the screen, written out."]]},{"start":280.75931249999996,"say":"Insert the inverse-square point-charge field. The area growth cancels the field's weakening, leaving q divided by epsilon zero.","live":null,"does":[[281.1073124999999,"point_work is shown on the screen, written out."],[287.00531249999995,"point_work is shown on the screen, written out."]]},{"start":290.24081249999995,"say":"Now deform the closed surface without moving it across the charge. Some patches tilt, some move closer, and others move farther away. The local dot products change, but the net signed crossing count cannot change.","live":null,"does":[[291.52931249999995,"shell is indicated — a transient flash."]]},{"start":304.77231249999994,"say":"Several charges simply add their fields. A charge inside contributes its full outward flux. A charge outside sends as much flux into the closed surface as it sends back out, so its net contribution is zero.","live":null,"does":[[308.58031249999993,"source is indicated — a transient flash."]]},{"start":319.28131249999996,"say":"The result is Gauss's law: the flux through any closed surface equals the net enclosed charge divided by epsilon zero. The surface may be imaginary, irregular, or placed wherever we choose.","live":null,"does":[[319.94231249999996,"point_result is shown on the screen, written out."],[331.64531249999993,"A box is drawn around point_result."],[332.55081249999995,"point_result is hidden from the screen — left the board."],[332.55081249999995,"point_work is hidden from the screen — left the board."],[332.55081249999995,"sphere_axes is hidden from the screen — left the board."],[332.55081249999995,"source is hidden from the screen — sphere_axes left the board."],[332.55081249999995,"shell is hidden from the screen — sphere_axes left the board."]]},{"start":333.75081249999994,"say":"Written in full, the closed-surface integral of E dot n hat equals enclosed charge divided by epsilon zero.","live":[],"does":[[333.75081249999994,"gauss_heading is shown on the screen, written out."],[336.29331249999996,"gauss_law is shown on the screen, written out."],[338.6503124999999,"gauss_definition is shown on the screen, written out."]]},{"start":342.05981249999996,"say":"Notice what the law does and does not say. It always gives total flux from enclosed charge. It gives the electric field itself only when symmetry makes the field's direction and magnitude simple on a carefully chosen surface.","live":["gauss_law","gauss_definition","gauss_heading"],"does":[[350.9643125,"gauss_law is indicated — a transient flash."]]},{"start":356.84631249999995,"say":"That choice is the real technique. For a spherical charge distribution we will choose a sphere. For a line we will choose a cylinder. For a sheet we will choose a pillbox. In every case, the source geometry chooses the useful Gaussian surface.","live":null,"does":[[372.0553124999999,"A box is drawn around gauss_law."],[373.12924999999996,"gauss_definition is hidden from the screen — left the board."],[373.12924999999996,"gauss_heading is hidden from the screen — left the board."],[373.12924999999996,"gauss_law is hidden from the screen — left the board."]]}]},{"title":"A Uniformly Charged Sphere","start":374.1709166666666,"end":586.4642083333332,"objects":{"axes":"an Axes3D (x_range=(-2.8, 2.8), y_range=(-2.8, 2.8), z_range=(-2.8, 2.8))","centre":"a Point [red] labelled \"O\" drawn in axes (location=(0.0, 0.0, 0.0))","charged_body":"a Sphere [red] drawn in axes (radius=1.2, opacity=0.24)","field_curve":"a FunctionPlot [blue] drawn in graph (function=<function>, x_range=(0.0, 3.0))","gaussian":"a Sphere [yellow] drawn in axes (radius=<VariableNumber gaussian_radius = 0.75>, opacity=0.11)","gaussian_radius":"a VariableNumber (initial_value=2.1)","graph":"an Axes (x_range=(0.0, 3.0), y_range=(0.0, 1.2), x_ticks_every=1.0)","graph_heading":"a Heading that says \"The Complete Field\"","inside_answer":"a Math [text] that says \"$E_(upright(\"in\"))(r) = frac(Q, 4 pi epsilon_0 R^3) r$\"","inside_heading":"a Heading that says \"Inside: $0 <= r <= R$\"","inside_label":"a Math [text] that says \"$frac(E, E_R) = frac(r, R)$\"","inside_work":"a Derivation [text] that says \"$rho = frac(Q, frac(4, 3) pi R^3) \\ q_(upright(\"enc\")) = rho frac(4, 3) pi r^3 \\ &= Q frac(r^3, R^3) \\ E(r) 4 pi r^2 = frac(Q r^3, epsilon_0 R^3)$\"","math":"a Math [text] that says \"$upright(\"total charge\") = Q, quad upright(\"radius\") = R$\"","outside_answer":"a Math [text] that says \"$E_(upright(\"out\"))(r) = frac(1, 4 pi epsilon_0) frac(Q, r^2)$\"","outside_heading":"a Heading that says \"Outside: $r >= R$\"","outside_label":"a Math [text] that says \"$frac(E, E_R) = frac(R^2, r^2)$\"","outside_work":"a Derivation [text] that says \"$Phi_E &= E(r) 4 pi r^2 \\ q_(upright(\"enc\")) &= Q \\ E(r) 4 pi r^2 &= frac(Q, epsilon_0)$\"","radial_arrows":"a Vector [blue] drawn in axes (start=(1.25, 0.0, 0.0), end=(2.15, 0.0, 0.0))","radial_arrows_2":"a Vector [blue] drawn in axes (start=(-1.25, 0.0, 0.0), end=(-2.15, 0.0, 0.0))","radial_arrows_3":"a Vector [blue] drawn in axes (start=(0.0, 1.25, 0.0), end=(0.0, 2.15, 0.0))","radial_arrows_4":"a Vector [blue] drawn in axes (start=(0.0, -1.25, 0.0), end=(0.0, -2.15, 0.0))","radial_arrows_5":"a Vector [blue] drawn in axes (start=(0.0, 0.0, 1.25), end=(0.0, 0.0, 2.15))","radial_arrows_6":"a Vector [blue] drawn in axes (start=(0.0, 0.0, -1.25), end=(0.0, 0.0, -2.15))","sphere_heading":"a Heading that says \"A Uniformly Charged Solid Sphere\"","surface_point":"a PlotPoint [yellow] labelled \"r=R\" drawn in graph (target='field_curve', x=1.0)"},"beats":[{"start":374.1709166666666,"say":"Take a solid insulating sphere of radius R, with total charge Q spread uniformly throughout its volume. Because the distribution looks the same after any rotation about its centre, the electric field must point radially.","live":[],"does":[[374.1709166666666,"sphere_heading is shown on the screen, written out."],[374.1709166666666,"axes is shown on the screen, written out."],[375.6569166666666,"charged_body is shown on the screen, faded in."],[384.5149166666666,"centre is shown on the screen, written out."],[386.82591666666656,"radial_arrows is shown on the screen, written out."],[386.82591666666656,"radial_arrows_2 is shown on the screen, written out."],[386.82591666666656,"radial_arrows_3 is shown on the screen, written out."],[386.82591666666656,"radial_arrows_4 is shown on the screen, written out."],[386.82591666666656,"radial_arrows_5 is shown on the screen, written out."],[386.82591666666656,"radial_arrows_6 is shown on the screen, written out."]]},{"start":388.3199166666666,"say":"A short turn confirms the geometry. There is no preferred direction around the sphere, so at a fixed distance from the centre every point must have the same field magnitude.","live":["axes","sphere_heading","charged_body","centre","radial_arrows","radial_arrows_2","radial_arrows_3","radial_arrows_4","radial_arrows_5","radial_arrows_6"],"does":[[388.3199166666666,"axes turns in its own slot."]]},{"start":399.8679166666666,"say":"First ask for the field outside the charge distribution. Choose a spherical Gaussian surface of radius r greater than R, centred on the same point.","live":null,"does":[[404.7209166666666,"gaussian is shown on the screen, faded in."],[410.01541666666657,"axes moves to a new place on the board."],[410.01541666666657,"sphere_heading is hidden from the screen — left the board."]]},{"start":410.6154166666666,"say":"On this Gaussian sphere, the field is everywhere normal to the surface and has one constant magnitude E of r. The flux is therefore E times four pi r squared.","live":["axes","charged_body","centre","radial_arrows","radial_arrows_2","radial_arrows_3","radial_arrows_4","radial_arrows_5","radial_arrows_6","gaussian"],"does":[[410.6154166666666,"outside_heading is shown on the screen, written out."],[418.21991666666656,"outside_work is shown on the screen, written out."]]},{"start":422.5699166666666,"say":"The Gaussian surface encloses the entire charged body, so the enclosed charge is Q.","live":["axes","charged_body","centre","radial_arrows","radial_arrows_2","radial_arrows_3","radial_arrows_4","radial_arrows_5","radial_arrows_6","gaussian","outside_heading"],"does":[[426.96991666666656,"outside_work is shown on the screen, written out."]]},{"start":429.3699166666666,"say":"Gauss's law sets the flux equal to Q over epsilon zero. Solving for E gives the familiar inverse-square field.","live":null,"does":[[429.7179166666666,"outside_work is shown on the screen, written out."],[434.2689166666666,"outside_answer is shown on the screen, written out."]]},{"start":438.4334166666666,"say":"Outside a spherically symmetric distribution, the field is exactly the same as if all the charge were concentrated at the centre. Gauss's law makes that statement exact, not merely approximate.","live":["axes","charged_body","centre","radial_arrows","radial_arrows_2","radial_arrows_3","radial_arrows_4","radial_arrows_5","radial_arrows_6","gaussian","outside_answer","outside_heading"],"does":[[442.7179166666666,"A box is drawn around outside_answer."],[446.79291666666654,"The box around outside_answer is lifted."],[451.4369166666666,"outside_answer is hidden from the screen — left the board."],[451.4369166666666,"outside_heading is hidden from the screen — left the board."],[451.4369166666666,"outside_work is hidden from the screen — left the board."]]},{"start":452.6369166666666,"say":"Now move the Gaussian sphere inside the charged body. Its radius r is less than R. The field still has spherical symmetry, but the surface now encloses only part of the total charge.","live":["axes","charged_body","centre","radial_arrows","radial_arrows_2","radial_arrows_3","radial_arrows_4","radial_arrows_5","radial_arrows_6","gaussian"],"does":[[452.6369166666666,"inside_heading is shown on the screen, written out."],[454.7029166666666,"gaussian is redrawn as the numbers it depends on change."],[454.7029166666666,"gaussian_radius ticks to 0.75."]]},{"start":465.4739166666666,"say":"Uniform volume charge density means total charge divided by total volume. The density rho is Q over four thirds pi R cubed.","live":["axes","charged_body","centre","radial_arrows","radial_arrows_2","radial_arrows_3","radial_arrows_4","radial_arrows_5","radial_arrows_6","gaussian","inside_heading"],"does":[[467.0409166666666,"inside_work is shown on the screen, written out."]]},{"start":475.0719166666666,"say":"The smaller Gaussian sphere has volume four thirds pi r cubed. Multiply that volume by rho to find the charge it encloses.","live":null,"does":[[477.3819166666666,"inside_work is shown on the screen, written out."]]},{"start":484.9484166666666,"say":"The common factors cancel, leaving Q times r cubed over R cubed. This fraction is simply the fraction of the charged volume lying inside the Gaussian surface.","live":null,"does":[[487.14291666666657,"inside_work is shown on the screen, written out."]]},{"start":496.70541666666657,"say":"Gauss's law again says E times four pi r squared equals enclosed charge over epsilon zero. One power of r survives after division.","live":null,"does":[[497.05391666666657,"inside_work is shown on the screen, written out."]]},{"start":508.7529166666666,"say":"The interior field is proportional to r. It is zero at the centre, where all directions balance, and it grows linearly as the Gaussian sphere encloses more charge.","live":null,"does":[[510.2739166666666,"inside_answer is shown on the screen, written out."],[516.1949166666666,"A box is drawn around inside_answer."],[519.9919166666666,"axes is hidden from the screen — left the board."],[519.9919166666666,"charged_body is hidden from the screen — axes left the board."],[519.9919166666666,"centre is hidden from the screen — axes left the board."],[519.9919166666666,"radial_arrows is hidden from the screen — axes left the board."],[519.9919166666666,"radial_arrows_2 is hidden from the screen — axes left the board."],[519.9919166666666,"radial_arrows_3 is hidden from the screen — axes left the board."],[519.9919166666666,"radial_arrows_4 is hidden from the screen — axes left the board."],[519.9919166666666,"radial_arrows_5 is hidden from the screen — axes left the board."],[519.9919166666666,"radial_arrows_6 is hidden from the screen — axes left the board."],[519.9919166666666,"gaussian is hidden from the screen — axes left the board."],[519.9919166666666,"inside_answer is hidden from the screen — left the board."],[519.9919166666666,"inside_heading is hidden from the screen — left the board."],[519.9919166666666,"inside_work is hidden from the screen — left the board."]]},{"start":521.1919166666665,"say":"Put the two regions on one graph. Distance is measured in units of R, and field strength in units of its value at the surface.","live":[],"does":[[521.1919166666665,"graph_heading is shown on the screen, written out."],[522.9449166666666,"graph is shown on the screen, written out."],[522.9449166666666,"field_curve is shown on the screen, written out."]]},{"start":530.3014166666666,"say":"Inside, the field rises in a straight line from zero. In normalized form, E over E at the surface equals r over R. Enclosed charge grows as r cubed, while Gaussian area grows as r squared, leaving one power of r.","live":["graph","graph_heading","field_curve"],"does":[[532.6349166666665,"field_curve is indicated — a transient flash."],[534.8869166666666,"graph moves to a new place on the board."],[534.8869166666666,"inside_label is shown on the screen, written out."]]},{"start":547.6544166666665,"say":"At r equals R, the inside and outside formulas agree. There is no jump in the field because the charge fills a volume rather than sitting in an infinitesimally thin surface layer.","live":["inside_label","graph","graph_heading","field_curve"],"does":[[548.3049166666665,"surface_point is shown on the screen, written out."]]},{"start":560.6424166666666,"say":"Outside, E over the surface field equals R squared over r squared. The enclosed charge has stopped growing, while the Gaussian area continues to grow as r squared.","live":["inside_label","graph","graph_heading","field_curve","surface_point"],"does":[[560.9909166666665,"outside_label is shown on the screen, written out."]]},{"start":573.3514166666666,"say":"The method was the same in both regions: use the source symmetry to choose a concentric sphere, make E constant on that surface, and then count only the charge actually enclosed.","live":["inside_label","outside_label","graph","graph_heading","field_curve","surface_point"],"does":[[574.7329166666666,"surface_point is indicated — a transient flash."],[585.4225416666666,"graph is hidden from the screen — left the board."],[585.4225416666666,"field_curve is hidden from the screen — graph left the board."],[585.4225416666666,"surface_point is hidden from the screen — graph left the board."],[585.4225416666666,"graph_heading is hidden from the screen — left the board."],[585.4225416666666,"inside_label is hidden from the screen — left the board."],[585.4225416666666,"outside_label is hidden from the screen — left the board."]]}]},{"title":"Lines, Sheets, and the Surfaces They Choose","start":586.4642083333332,"end":816.7534166666665,"objects":{"charged_line":"a Line [red] labelled \"lambda\" drawn in line_axes (start=(0.0, 0.0, -2.4), end=(0.0, 0.0, 2.4))","charged_sheet":"a Plane [red] labelled \"sigma\" drawn in sheet_axes (size=3.7, opacity=0.22)","comparison_heading":"a Heading that says \"Geometry Controls the Distance Law\"","line_answer":"a Math [text] that says \"$E_(upright(\"line\"))(r) = frac(lambda, 2 pi epsilon_0 r)$\"","line_arrows":"a Vector [blue] drawn in line_axes (start=(0.15, 0.0, -0.8), end=(1.7, 0.0, -0.8))","line_arrows_2":"a Vector [blue] drawn in line_axes (start=(-0.15, 0.0, -0.8), end=(-1.7, 0.0, -0.8))","line_arrows_3":"a Vector [blue] drawn in line_axes (start=(0.0, 0.15, 0.0), end=(0.0, 1.7, 0.0))","line_arrows_4":"a Vector [blue] drawn in line_axes (start=(0.0, -0.15, 0.0), end=(0.0, -1.7, 0.0))","line_arrows_5":"a Vector [blue] drawn in line_axes (start=(0.15, 0.0, 0.8), end=(1.7, 0.0, 0.8))","line_arrows_6":"a Vector [blue] drawn in line_axes (start=(-0.15, 0.0, 0.8), end=(-1.7, 0.0, 0.8))","line_axes":"an Axes3D (x_range=(-2.4, 2.4), y_range=(-2.4, 2.4), z_range=(-2.6, 2.6))","line_caption":"a Text [text] that says \"Line: area per length grows like $r$.\"","line_case":"a Math [text] that says \"$E(r) = frac(C, r)$\"","line_cylinder":"a Cylinder [yellow] drawn in line_axes (start=(0.0, 0.0, -1.35), end=(0.0, 0.0, 1.35), fill_end_circles=False)","line_heading":"a Heading that says \"An Infinite Uniform Line Charge\"","line_work":"a Derivation [text] that says \"$Phi_E &= E (2 pi r L) \\ q_(upright(\"enc\")) &= lambda L \\ E (2 pi r L) &= frac(lambda L, epsilon_0)$\"","pillbox":"a Cylinder [yellow] drawn in sheet_axes (start=(0.0, 0.0, -0.85), end=(0.0, 0.0, 0.85), fill_end_circles=False)","point_caption":"a Text [text] that says \"Point: area grows like $r^2$.\"","point_case":"a Math [text] that says \"$E(r) = frac(C, r^2)$\"","sheet_answer":"a Math [text] that says \"$E_(upright(\"sheet\")) = frac(sigma, 2 epsilon_0)$\"","sheet_arrows":"a Vector [blue] drawn in sheet_axes (start=(-1.0, -0.5, 0.1), end=(-1.0, -0.5, 1.65))","sheet_arrows_2":"a Vector [blue] drawn in sheet_axes (start=(0.0, 0.0, 0.1), end=(0.0, 0.0, 1.65))","sheet_arrows_3":"a Vector [blue] drawn in sheet_axes (start=(1.0, 0.5, 0.1), end=(1.0, 0.5, 1.65))","sheet_arrows_4":"a Vector [blue] drawn in sheet_axes (start=(-1.0, -0.5, -0.1), end=(-1.0, -0.5, -1.65))","sheet_arrows_5":"a Vector [blue] drawn in sheet_axes (start=(0.0, 0.0, -0.1), end=(0.0, 0.0, -1.65))","sheet_arrows_6":"a Vector [blue] drawn in sheet_axes (start=(1.0, 0.5, -0.1), end=(1.0, 0.5, -1.65))","sheet_axes":"an Axes3D (x_range=(-2.4, 2.4), y_range=(-2.4, 2.4), z_range=(-2.2, 2.2))","sheet_caption":"a Text [text] that says \"Sheet: the two cap areas stay fixed.\"","sheet_case":"a Math [text] that says \"$E(r) = C$\"","sheet_heading":"a Heading that says \"An Infinite Uniform Sheet\"","sheet_work":"a Derivation [text] that says \"$Phi_E &= E A + E A = 2 E A \\ q_(upright(\"enc\")) &= sigma A \\ 2 E A &= frac(sigma A, epsilon_0)$\""},"beats":[{"start":586.4642083333332,"say":"Now stretch the charge distribution into an ideal infinite line with uniform charge per length lambda. Translation along the line changes nothing, and rotation around it changes nothing.","live":[],"does":[[586.4642083333332,"line_heading is shown on the screen, written out."],[586.4642083333332,"line_axes is shown on the screen, written out."],[589.8432083333332,"charged_line is shown on the screen, written out."]]},{"start":599.0107083333332,"say":"Those symmetries force the electric field to point directly away from the line. Its magnitude can depend on perpendicular distance r, but not on position along the line or angle around it.","live":["line_axes","line_heading","charged_line"],"does":[[602.6102083333332,"line_arrows is shown on the screen, written out."],[602.6102083333332,"line_arrows_2 is shown on the screen, written out."],[602.6102083333332,"line_arrows_3 is shown on the screen, written out."],[602.6102083333332,"line_arrows_4 is shown on the screen, written out."],[602.6102083333332,"line_arrows_5 is shown on the screen, written out."],[602.6102083333332,"line_arrows_6 is shown on the screen, written out."]]},{"start":612.1032083333332,"say":"Choose a cylinder centred on the line. Its curved side is everywhere the same distance r from the charge, so E has one constant magnitude there.","live":["line_axes","line_heading","charged_line","line_arrows","line_arrows_2","line_arrows_3","line_arrows_4","line_arrows_5","line_arrows_6"],"does":[[612.8582083333332,"line_cylinder is shown on the screen, faded in."]]},{"start":622.1537083333332,"say":"Let the solid settle into view. The charged line is the cylinder's axis, and the blue field arrows point through its curved wall.","live":["line_axes","line_heading","charged_line","line_arrows","line_arrows_2","line_arrows_3","line_arrows_4","line_arrows_5","line_arrows_6","line_cylinder"],"does":[[622.1537083333332,"line_axes turns in its own slot."]]},{"start":631.1017083333332,"say":"On the two end caps, the outward normals point along the line, while the electric field points radially away from it. Their dot product is zero, so the caps contribute no flux.","live":null,"does":[[631.8912083333332,"line_cylinder is indicated — a transient flash."]]},{"start":643.2192083333332,"say":"Only the curved side contributes. Its area is circumference two pi r times length L, so the flux is E times two pi r L.","live":null,"does":[[650.5332083333332,"line_axes moves to a new place on the board."],[650.5332083333332,"line_work is shown on the screen, written out."]]},{"start":654.1572083333332,"say":"The cylinder encloses a length L of line charge. Charge per length lambda times L gives enclosed charge lambda L.","live":null,"does":[[660.5952083333332,"line_work is shown on the screen, written out."]]},{"start":663.3077083333332,"say":"Apply Gauss's law. The length L appears on both sides and cancels. Solving leaves lambda over two pi epsilon zero r.","live":null,"does":[[664.0622083333332,"line_work is shown on the screen, written out."],[669.5072083333332,"line_answer is shown on the screen, written out."]]},{"start":673.3232083333332,"say":"The line field falls as one over r, not one over r squared. As the cylinder expands, its relevant area per unit length grows only in proportion to r.","live":["line_answer","line_axes","line_heading","charged_line","line_arrows","line_arrows_2","line_arrows_3","line_arrows_4","line_arrows_5","line_arrows_6","line_cylinder"],"does":[[674.8902083333332,"A box is drawn around line_answer."],[684.0622083333332,"line_answer is hidden from the screen — left the board."],[684.0622083333332,"line_axes is hidden from the screen — left the board."],[684.0622083333332,"charged_line is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_arrows is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_arrows_2 is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_arrows_3 is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_arrows_4 is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_arrows_5 is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_arrows_6 is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_cylinder is hidden from the screen — line_axes left the board."],[684.0622083333332,"line_heading is hidden from the screen — left the board."],[684.0622083333332,"line_work is hidden from the screen — left the board."]]},{"start":685.2622083333332,"say":"Next spread charge uniformly across an ideal infinite sheet, with surface charge density sigma. Sliding anywhere within the sheet cannot change the field.","live":[],"does":[[685.2622083333332,"sheet_heading is shown on the screen, written out."],[685.2622083333332,"sheet_axes is shown on the screen, written out."],[689.1512083333332,"charged_sheet is shown on the screen, faded in."]]},{"start":696.4387083333332,"say":"Rotating the sheet within its own plane also changes nothing. The only distinguished direction is perpendicular to the sheet, so the field must point normally away on both sides.","live":["sheet_axes","sheet_heading","charged_sheet"],"does":[[702.8242083333332,"sheet_arrows is shown on the screen, written out."],[702.8242083333332,"sheet_arrows_2 is shown on the screen, written out."],[702.8242083333332,"sheet_arrows_3 is shown on the screen, written out."],[702.8242083333332,"sheet_arrows_4 is shown on the screen, written out."],[702.8242083333332,"sheet_arrows_5 is shown on the screen, written out."],[702.8242083333332,"sheet_arrows_6 is shown on the screen, written out."]]},{"start":708.1497083333333,"say":"Choose a short cylindrical pillbox that straddles the sheet. Its flat caps are parallel to the charge distribution, and its curved wall joins them.","live":["sheet_axes","sheet_heading","charged_sheet","sheet_arrows","sheet_arrows_2","sheet_arrows_3","sheet_arrows_4","sheet_arrows_5","sheet_arrows_6"],"does":[[709.8102083333332,"pillbox is shown on the screen, faded in."]]},{"start":718.2927083333332,"say":"A small turn shows the construction. The field passes through the two caps, while it runs parallel to the curved wall.","live":["sheet_axes","sheet_heading","charged_sheet","sheet_arrows","sheet_arrows_2","sheet_arrows_3","sheet_arrows_4","sheet_arrows_5","sheet_arrows_6","pillbox"],"does":[[718.2927083333332,"sheet_axes turns in its own slot."]]},{"start":726.7872083333332,"say":"The curved wall contributes zero flux because its normal lies within the sheet while E is perpendicular to it. Each cap contributes E A, so the total flux is two E A.","live":null,"does":[[737.3412083333332,"sheet_axes moves to a new place on the board."],[737.3412083333332,"sheet_work is shown on the screen, written out."]]},{"start":739.1132083333332,"say":"The pillbox encloses sheet area A, so it encloses charge sigma A.","live":null,"does":[[740.1472083333332,"sheet_work is shown on the screen, written out."]]},{"start":744.6422083333332,"say":"Gauss's law gives two E A equals sigma A over epsilon zero. The cap area cancels, leaving E equals sigma over two epsilon zero.","live":null,"does":[[744.9732083333331,"sheet_work is shown on the screen, written out."],[751.3932083333332,"sheet_answer is shown on the screen, written out."]]},{"start":755.1332083333332,"say":"There is no distance in the answer. For an ideal infinite sheet, moving the caps farther away does not spread a fixed bundle over a growing area. The same cap area intercepts the same flux.","live":["sheet_answer","sheet_axes","sheet_heading","charged_sheet","sheet_arrows","sheet_arrows_2","sheet_arrows_3","sheet_arrows_4","sheet_arrows_5","sheet_arrows_6","pillbox"],"does":[[755.8592083333332,"A box is drawn around sheet_answer."],[768.0262083333332,"sheet_answer is hidden from the screen — left the board."],[768.0262083333332,"sheet_axes is hidden from the screen — left the board."],[768.0262083333332,"charged_sheet is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_arrows is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_arrows_2 is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_arrows_3 is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_arrows_4 is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_arrows_5 is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_arrows_6 is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"pillbox is hidden from the screen — sheet_axes left the board."],[768.0262083333332,"sheet_heading is hidden from the screen — left the board."],[768.0262083333332,"sheet_work is hidden from the screen — left the board."]]},{"start":769.2262083333331,"say":"Put the three geometries together. A point spreads flux over a sphere whose area grows as r squared, so its field has the form a constant over r squared.","live":[],"does":[[769.2262083333331,"comparison_heading is shown on the screen, written out."],[772.4192083333331,"point_case is shown on the screen, written out."],[774.0092083333332,"point_caption is shown on the screen, written out."]]},{"start":781.0882083333331,"say":"A line spreads flux over the curved wall of a cylinder. Per unit length, that area grows as r, so the field has the form a constant over r.","live":["point_case","point_caption","comparison_heading"],"does":[[781.7382083333332,"line_case is shown on the screen, written out."],[784.2922083333332,"line_caption is shown on the screen, written out."]]},{"start":792.4152083333331,"say":"A sheet sends flux through two equal caps. Their area does not change when the pillbox grows taller, so the ideal sheet field is constant.","live":["point_case","point_caption","line_case","line_caption","comparison_heading"],"does":[[792.9262083333331,"sheet_case is shown on the screen, written out."],[794.8652083333332,"sheet_caption is shown on the screen, written out."]]},{"start":802.8722083333332,"say":"Gauss's law was identical in all three cases. What changed was the symmetry, and symmetry determined the surface on which E became constant and the unwanted pieces contributed zero.","live":["point_case","point_caption","line_case","line_caption","sheet_case","sheet_caption","comparison_heading"],"does":[[808.0842083333331,"point_case is indicated — a transient flash."],[808.2842083333331,"line_case is indicated — a transient flash."],[808.4842083333331,"sheet_case is indicated — a transient flash."],[815.7117499999998,"comparison_heading is hidden from the screen — left the board."],[815.7117499999998,"line_caption is hidden from the screen — left the board."],[815.7117499999998,"line_case is hidden from the screen — left the board."],[815.7117499999998,"point_caption is hidden from the screen — left the board."],[815.7117499999998,"point_case is hidden from the screen — left the board."],[815.7117499999998,"sheet_caption is hidden from the screen — left the board."],[815.7117499999998,"sheet_case is hidden from the screen — left the board."]]}]},{"title":"Conductors, Surface Charge, and the Car","start":816.7534166666665,"end":1028.8116041666665,"objects":{"body":"a Polygon [blue] labelled \"upright(\"metal body\")\" drawn in car (vertices=((0.7, 1.0), (1.0, 2.3), (2.1, 2.6), (3.0, 3.6), (5.2, 3.6), (6…, fill_opacity=0.22)","bolt_one":"a Line [yellow] drawn in car (start=(4.7, 5.1), end=(4.25, 4.45))","bolt_three":"a Line [yellow] drawn in car (start=(4.65, 4.0), end=(4.25, 3.6))","bolt_two":"a Line [yellow] drawn in car (start=(4.25, 4.45), end=(4.65, 4.0))","cabin":"a Polygon [gray] labelled \"upright(\"passenger space\")\" drawn in car (vertices=((2.2, 2.5), (3.1, 3.4), (5.0, 3.4), (5.9, 2.5)), filled=False, dashed=True)","cabin_field":"a Math [text] that says \"$arrow(E) approx arrow(0) quad upright(\"in the protected interior\")$\"","car":"a Figure (x_range=(0.0, 8.0), y_range=(0.0, 5.2), aspect=(8.0, 5.2))","car_heading":"a Heading that says \"Why a Metal Car Protects Its Interior\"","conductor_heading":"a Heading that says \"Electrostatic Equilibrium in a Conductor\"","conductor_work":"a Derivation [text] that says \"$Phi_E = integral.double_S arrow(E) dot hat(n) thin dif A \\ &= 0 \\ q_(upright(\"enc\")) = epsilon_0 Phi_E = 0$\"","equilibrium_note":"a Panel that says \"In electrostatic equilibrium, mobile charge has finished rearranging. The electric field inside the conducting material is zero.\"","inner_gaussian":"a Circle [gray] labelled \"S\" drawn in metal_picture (radius=0.72)","left_current":"a Vector [red] labelled \"upright(\"surface current\")\" drawn in car (start=(4.2, 3.55), end=(1.15, 2.15))","metal":"a Circle [blue] drawn in metal_picture (radius=1.35, filled=True)","metal_picture":"a Figure (x_range=(-2.7, 2.7), y_range=(-2.3, 2.3), aspect=(5.4, 4.6))","outer_fields":"a Vector [yellow] drawn in metal_picture (start=(1.42, 0.0), end=(2.25, 0.0))","outer_fields_2":"a Vector [yellow] drawn in metal_picture (start=(-1.42, 0.0), end=(-2.25, 0.0))","outer_fields_3":"a Vector [yellow] drawn in metal_picture (start=(0.0, 1.42), end=(0.0, 2.1))","outer_fields_4":"a Vector [yellow] drawn in metal_picture (start=(0.0, -1.42), end=(0.0, -2.1))","outer_fields_5":"a Vector [yellow] drawn in metal_picture (start=(1.0, 1.0), end=(1.62, 1.62))","outer_fields_6":"a Vector [yellow] drawn in metal_picture (start=(-1.0, 1.0), end=(-1.62, 1.62))","outer_fields_7":"a Vector [yellow] drawn in metal_picture (start=(1.0, -1.0), end=(1.62, -1.62))","outer_fields_8":"a Vector [yellow] drawn in metal_picture (start=(-1.0, -1.0), end=(-1.62, -1.62))","passenger_left":"a Point [green] drawn in car (location=(3.5, 2.65))","passenger_right":"a Point [green] drawn in car (location=(4.55, 2.65))","recap":"a Block [text] that says \"Flux through a closed surface counts enclosed charge. Symmetry chooses a surface on which the flux is easy to calculate. Mobile charge in a conductor rearranges until the field in the metal is zero.\"","recap_heading":"a Heading that says \"What to Carry Away\"","right_current":"a Vector [red] drawn in car (start=(4.35, 3.55), end=(6.85, 2.1))","safety_note":"a Text [text] that says \"The shell redirects charge and current around the passenger compartment. The protection comes from the metal body, not from the tires.\"","surface_charges":"a Point [red] drawn in metal_picture (location=(1.35, 0.0))","surface_charges_2":"a Point [red] drawn in metal_picture (location=(1.0341599982106204, 0.8677632730768281))","surface_charges_3":"a Point [red] drawn in metal_picture (location=(0.23442503985035607, 1.3294904665664808))","surface_charges_4":"a Point [red] drawn in metal_picture (location=(-0.6749999999999997, 1.1691342951089922))","surface_charges_5":"a Point [red] drawn in metal_picture (location=(-1.2685850380609762, 0.461727193489653))","surface_charges_6":"a Point [red] drawn in metal_picture (location=(-1.2685850380609764, -0.46172719348965274))","surface_charges_7":"a Point [red] drawn in metal_picture (location=(-0.6750000000000006, -1.1691342951089918))","surface_charges_8":"a Point [red] drawn in metal_picture (location=(0.23442503985035548, -1.329490466566481))","surface_charges_9":"a Point [red] drawn in metal_picture (location=(1.03415999821062, -0.8677632730768285))","surface_heading":"a Heading that says \"Where the Excess Charge Goes\"","surface_result":"a Math [text] that says \"$q_(upright(\"bulk\")) = 0$\"","wheel_left":"a Circle [gray] drawn in car (center=(2.0, 0.85), radius=0.55)","wheel_right":"a Circle [gray] drawn in car (center=(6.0, 0.85), radius=0.55)","zero_field":"a Math [text] that says \"$arrow(E) = arrow(0) quad upright(\"inside conducting material\")$\""},"beats":[{"start":816.7534166666665,"say":"A conductor contains mobile charge. If an electric field existed inside the metal, that charge would feel a force and begin to move. A state with moving charge is not electrostatic equilibrium.","live":[],"does":[[816.7534166666665,"conductor_heading is shown on the screen, written out."],[816.9164166666665,"metal_picture is shown on the screen, written out."],[821.2934166666664,"metal is shown on the screen, faded in."]]},{"start":829.4279166666664,"say":"The mobile charge rearranges on an extremely short time scale. Its own electric field opposes the field that drove the motion, and equilibrium is reached only when the net electric field inside the conducting material is zero.","live":["metal_picture","conductor_heading","metal"],"does":[[837.9374166666664,"metal_picture moves to a new place on the board."],[837.9374166666664,"equilibrium_note is shown on the screen, written out."],[842.2104166666664,"zero_field is shown on the screen, written out."]]},{"start":843.6694166666665,"say":"That statement comes from the physics of mobile charge, not from Gauss's law alone. Gauss's law now tells us an important consequence of the zero field.","live":["equilibrium_note","zero_field","metal_picture","conductor_heading","metal"],"does":[[852.1674166666664,"zero_field is indicated — a transient flash."],[853.3514166666664,"metal_picture moves to a new place on the board."],[853.3514166666664,"conductor_heading is hidden from the screen — left the board."],[853.3514166666664,"equilibrium_note is hidden from the screen — left the board."],[853.3514166666664,"zero_field is hidden from the screen — left the board."]]},{"start":853.9514166666664,"say":"Draw any closed Gaussian surface lying completely inside the conducting material. Because E is zero at every point on it, its electric flux is zero.","live":["metal_picture","metal"],"does":[[853.9514166666664,"surface_heading is shown on the screen, written out."],[855.2984166666664,"inner_gaussian is shown on the screen, written out."],[860.1634166666664,"conductor_work is shown on the screen, written out."],[862.5784166666665,"conductor_work is shown on the screen, written out."]]},{"start":864.5479166666664,"say":"Gauss's law then says that the net charge enclosed by every such interior surface is zero.","live":["metal_picture","metal","surface_heading","inner_gaussian"],"does":[[867.0324166666664,"conductor_work is shown on the screen, written out."]]},{"start":871.0109166666665,"say":"Excess charge therefore cannot remain distributed through the bulk of the metal in electrostatic equilibrium. It moves outward and comes to rest on the conductor's surface.","live":null,"does":[[874.0874166666664,"surface_result is shown on the screen, written out."],[880.1594166666664,"surface_charges is shown on the screen, written out."],[880.1594166666664,"surface_charges_2 is shown on the screen, written out."],[880.1594166666664,"surface_charges_3 is shown on the screen, written out."],[880.1594166666664,"surface_charges_4 is shown on the screen, written out."],[880.1594166666664,"surface_charges_5 is shown on the screen, written out."],[880.1594166666664,"surface_charges_6 is shown on the screen, written out."],[880.1594166666664,"surface_charges_7 is shown on the screen, written out."],[880.1594166666664,"surface_charges_8 is shown on the screen, written out."],[880.1594166666664,"surface_charges_9 is shown on the screen, written out."]]},{"start":881.7234166666665,"say":"Outside the conductor, those surface charges can produce an electric field. Inside the conducting material, they have arranged themselves so that their combined field cancels.","live":["metal_picture","metal","surface_result","surface_heading","inner_gaussian","surface_charges","surface_charges_2","surface_charges_3","surface_charges_4","surface_charges_5","surface_charges_6","surface_charges_7","surface_charges_8","surface_charges_9"],"does":[[882.0714166666664,"outer_fields is shown on the screen, written out."],[882.0714166666664,"outer_fields_2 is shown on the screen, written out."],[882.0714166666664,"outer_fields_3 is shown on the screen, written out."],[882.0714166666664,"outer_fields_4 is shown on the screen, written out."],[882.0714166666664,"outer_fields_5 is shown on the screen, written out."],[882.0714166666664,"outer_fields_6 is shown on the screen, written out."],[882.0714166666664,"outer_fields_7 is shown on the screen, written out."],[882.0714166666664,"outer_fields_8 is shown on the screen, written out."]]},{"start":892.9464166666664,"say":"The surface distribution need not be uniform. Charge crowds more strongly near sharp points, where the surrounding field can become especially large. But the equilibrium condition inside the metal remains E equals zero.","live":["metal_picture","metal","surface_result","surface_heading","inner_gaussian","surface_charges","surface_charges_2","surface_charges_3","surface_charges_4","surface_charges_5","surface_charges_6","surface_charges_7","surface_charges_8","surface_charges_9","outer_fields","outer_fields_2","outer_fields_3","outer_fields_4","outer_fields_5","outer_fields_6","outer_fields_7","outer_fields_8"],"does":[[905.2414166666665,"A box is drawn around surface_result."],[906.0654166666665,"conductor_work is hidden from the screen — left the board."],[906.0654166666665,"metal_picture is hidden from the screen — left the board."],[906.0654166666665,"metal is hidden from the screen — metal_picture left the board."],[906.0654166666665,"inner_gaussian is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_2 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_3 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_4 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_5 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_6 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_7 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_8 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_charges_9 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_2 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_3 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_4 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_5 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_6 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_7 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"outer_fields_8 is hidden from the screen — metal_picture left the board."],[906.0654166666665,"surface_heading is hidden from the screen — left the board."],[906.0654166666665,"surface_result is hidden from the screen — left the board."]]},{"start":907.2654166666664,"say":"Now replace the simple conductor with the metal body of a hard-top car. The body forms a conducting shell around the passenger compartment.","live":[],"does":[[907.2654166666664,"car_heading is shown on the screen, written out."],[907.2654166666664,"car is shown on the screen, written out."],[909.9124166666664,"body is shown on the screen, faded in."],[914.1614166666665,"cabin is shown on the screen, written out."],[914.1614166666665,"wheel_left is shown on the screen, written out."],[914.1614166666665,"wheel_right is shown on the screen, written out."],[914.1614166666665,"passenger_left is shown on the screen, written out."],[914.1614166666665,"passenger_right is shown on the screen, written out."]]},{"start":916.2244166666665,"say":"Suppose lightning strikes the roof. The lightning delivers charge and a large current to the exterior metal.","live":["car","car_heading","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right"],"does":[[917.0604166666665,"bolt_one is shown on the screen, written out."],[917.0604166666665,"bolt_two is shown on the screen, written out."],[917.0604166666665,"bolt_three is shown on the screen, written out."]]},{"start":923.7679166666665,"say":"The charge spreads over the outside, and the current finds conducting paths along the exterior body. The metal shell carries the dangerous electrical disturbance around the passenger space rather than through it.","live":["car","car_heading","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right","bolt_one","bolt_two","bolt_three"],"does":[[925.5204166666665,"left_current is shown on the screen, written out."],[925.5204166666665,"right_current is shown on the screen, written out."]]},{"start":936.6169166666665,"say":"The passenger compartment is therefore close to one electric potential, with a strongly reduced electric field inside. This shielding behavior is often called the Faraday-cage effect.","live":["car","car_heading","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right","bolt_one","bolt_two","bolt_three","left_current","right_current"],"does":[[941.4004166666665,"cabin_field is shown on the screen, written out."],[944.0584166666665,"cabin is indicated — a transient flash."]]},{"start":948.0489166666664,"say":"The protection comes from the continuous metal shell, not primarily from the rubber tires. During a storm, occupants should remain inside with windows closed and avoid touching metal parts connected to the exterior.","live":["car","cabin_field","car_heading","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right","bolt_one","bolt_two","bolt_three","left_current","right_current"],"does":[[948.5604166666665,"safety_note is shown on the screen, written out."]]},{"start":961.6639166666664,"say":"The same principle is used deliberately in shielded rooms, cable coverings, and metal enclosures around sensitive electronics. Conductors rearrange charge so that their protected interiors experience very little electric field.","live":["car","cabin_field","safety_note","car_heading","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right","bolt_one","bolt_two","bolt_three","left_current","right_current"],"does":[[972.2294166666665,"cabin_field is indicated — a transient flash."],[975.1549166666664,"cabin_field moves to a new place on the board."],[975.1549166666664,"car moves to a new place on the board."],[975.1549166666664,"car_heading is hidden from the screen — left the board."],[975.1549166666664,"safety_note is hidden from the screen — left the board."],[975.1549166666664,"recap is shown on the screen, written out."]]},{"start":976.3549166666664,"say":"Three ideas organize the whole lecture. First, electric flux is the signed amount of electric field crossing a surface, and the flux through a closed surface counts net enclosed charge.","live":["car","cabin_field","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right","bolt_one","bolt_two","bolt_three","left_current","right_current","recap"],"does":[[976.3549166666664,"recap_heading is shown on the screen, written out."],[979.4894166666664,"recap (the \"Flux through a closed surface counts enclosed charge.\" part) is indicated — a transient flash."]]},{"start":987.7749166666665,"say":"Second, Gauss's law becomes a field-solving tool only when symmetry chooses a useful surface: a sphere for spherical charge, a cylinder for a line, and a pillbox for a sheet.","live":["car","cabin_field","body","cabin","wheel_left","wheel_right","passenger_left","passenger_right","bolt_one","bolt_two","bolt_three","left_current","right_current","recap","recap_heading"],"does":[[988.0764166666664,"recap (the \"Symmetry chooses a surface on which the flux is easy to calculate.\" part) is indicated — a transient flash."]]},{"start":1000.3439166666665,"say":"Third, mobile charge in a conductor rearranges until the field inside the metal is zero. Excess charge lives on the surface, and a closed metal body redirects an external electrical disturbance around its interior.","live":null,"does":[[1000.9014166666664,"recap (the \"Mobile charge in a conductor rearranges until the field in the metal is zero.\" part) is indicated — a transient flash."]]},{"start":1014.0869166666664,"say":"Flux made the counting picture precise. Gauss's law connected that count to charge. Symmetry turned the law into three electric fields, and electrostatic equilibrium turned it into protection inside a conductor.","live":null,"does":[[1025.7554166666664,"A box is drawn around cabin_field."],[1027.7699374999997,"cabin_field is hidden from the screen — left the board."],[1027.7699374999997,"car is hidden from the screen — left the board."],[1027.7699374999997,"body is hidden from the screen — car left the board."],[1027.7699374999997,"cabin is hidden from the screen — car left the board."],[1027.7699374999997,"wheel_left is hidden from the screen — car left the board."],[1027.7699374999997,"wheel_right is hidden from the screen — car left the board."],[1027.7699374999997,"passenger_left is hidden from the screen — car left the board."],[1027.7699374999997,"passenger_right is hidden from the screen — car left the board."],[1027.7699374999997,"bolt_one is hidden from the screen — car left the board."],[1027.7699374999997,"bolt_two is hidden from the screen — car left the board."],[1027.7699374999997,"bolt_three is hidden from the screen — car left the board."],[1027.7699374999997,"left_current is hidden from the screen — car left the board."],[1027.7699374999997,"right_current is hidden from the screen — car left the board."],[1027.7699374999997,"recap is hidden from the screen — left the board."],[1027.7699374999997,"recap_heading is hidden from the screen — left the board."]]}]}]},"durationSeconds":1029,"chapters":[{"title":"Flux That Refuses to Change","startSeconds":0,"narration":"Place one positive point charge in empty space. At every surrounding point, its electric field points directly away from the charge. These blue arrows represent field lines, a drawing that lets us follow the direction of the field through space. The picture is three dimensional. Every direction away from the charge is equivalent, and no radial direction is preferred over another. Now surround the charge with a small sphere. Every blue line leaving the charge crosses that sphere once, from its inside to its outside. If we use the lines as a counting picture, the outward flux is the number of crossings. The lines are not physical threads, and their drawn number is arbitrary. Still, the crossing picture captures something real: a surface facing the field receives positive flux, and a closed surface surrounding the source intercepts the whole outward pattern. Replace the small sphere with a larger one. The same radial lines cross it, so the crossing count has not changed. They are farther apart now, which is the picture's way of saying that the electric field is weaker. The radius grew, and the sphere's area grew as radius squared. But the field of a point charge weakened as one over radius squared. Twice the radius gives four times the area and one quarter of the field strength. Their product stays fixed. A sphere is not essential. Dent the surface here, bulge it there, and keep the charge enclosed. Every ray that leaves the charge still has to cross the closed boundary. The local crossing angles and local field strengths change, but the total outward count does not. For a sphere of radius r, the point-charge field has magnitude one over four pi epsilon zero, times q over r squared. The sphere's area is four pi r squared. This is exactly the geometrical growth that compensates for the inverse-square weakening. Multiply field strength by area. The r squared in the area cancels the r squared in the denominator, and four pi cancels as well. What remains is q divided by epsilon zero. It has no radius in it. The calculation therefore agrees with the crossing picture: every surrounding sphere receives the same total outward electric flux. The lumpy surface needs a more careful sum because its field strength and angle vary from patch to patch. But once those local contributions are counted correctly, the same enclosed charge gives the same total flux. That invariant total is the idea Gauss's law will make exact. The next step is to define what one small crossing contributes, including the angle at which the field meets the surface."},{"title":"From Crossings to Gauss's Law","startSeconds":177.10831249999995,"narration":"Zoom in on one small piece of a surface. Its area is delta A, and the yellow arrow is the outward unit normal, the direction perpendicular to that patch. Let the electric field meet the patch at an angle theta from the normal. The field crosses most effectively when it points along the normal. The patch's flux is E dot n hat times delta A. Equivalently, it is field strength times area times cosine theta. The cosine selects the normal component of the field. If the field runs parallel to the patch, theta is ninety degrees and the cosine is zero. Lines may skim along the surface, but none cross it, so that part contributes no flux. If the field points outward, the dot product is positive. If it points inward, the dot product is negative. Flux therefore counts signed crossings, outward minus inward. A curved surface is built from many small patches. Add the local dot products over all of them. As the patches become arbitrarily small, the sum becomes a surface integral. This is the precise definition of electric flux. Unlike the number of lines an artist draws, the integral has a fixed numerical meaning. Return to a point charge and a spherical surface around it. Symmetry makes the field radial, so it points along the outward normal everywhere on the sphere. Symmetry also makes the field strength the same at every point of a sphere with radius r. We can therefore take E outside the integral, leaving the total area four pi r squared. Insert the inverse-square point-charge field. The area growth cancels the field's weakening, leaving q divided by epsilon zero. Now deform the closed surface without moving it across the charge. Some patches tilt, some move closer, and others move farther away. The local dot products change, but the net signed crossing count cannot change. Several charges simply add their fields. A charge inside contributes its full outward flux. A charge outside sends as much flux into the closed surface as it sends back out, so its net contribution is zero. The result is Gauss's law: the flux through any closed surface equals the net enclosed charge divided by epsilon zero. The surface may be imaginary, irregular, or placed wherever we choose. Written in full, the closed-surface integral of E dot n hat equals enclosed charge divided by epsilon zero. Notice what the law does and does not say. It always gives total flux from enclosed charge. It gives the electric field itself only when symmetry makes the field's direction and magnitude simple on a carefully chosen surface. That choice is the real technique. For a spherical charge distribution we will choose a sphere. For a line we will choose a cylinder. For a sheet we will choose a pillbox. In every case, the source geometry chooses the useful Gaussian surface."},{"title":"A Uniformly Charged Sphere","startSeconds":374.1709166666666,"narration":"Take a solid insulating sphere of radius R, with total charge Q spread uniformly throughout its volume. Because the distribution looks the same after any rotation about its centre, the electric field must point radially. A short turn confirms the geometry. There is no preferred direction around the sphere, so at a fixed distance from the centre every point must have the same field magnitude. First ask for the field outside the charge distribution. Choose a spherical Gaussian surface of radius r greater than R, centred on the same point. On this Gaussian sphere, the field is everywhere normal to the surface and has one constant magnitude E of r. The flux is therefore E times four pi r squared. The Gaussian surface encloses the entire charged body, so the enclosed charge is Q. Gauss's law sets the flux equal to Q over epsilon zero. Solving for E gives the familiar inverse-square field. Outside a spherically symmetric distribution, the field is exactly the same as if all the charge were concentrated at the centre. Gauss's law makes that statement exact, not merely approximate. Now move the Gaussian sphere inside the charged body. Its radius r is less than R. The field still has spherical symmetry, but the surface now encloses only part of the total charge. Uniform volume charge density means total charge divided by total volume. The density rho is Q over four thirds pi R cubed. The smaller Gaussian sphere has volume four thirds pi r cubed. Multiply that volume by rho to find the charge it encloses. The common factors cancel, leaving Q times r cubed over R cubed. This fraction is simply the fraction of the charged volume lying inside the Gaussian surface. Gauss's law again says E times four pi r squared equals enclosed charge over epsilon zero. One power of r survives after division. The interior field is proportional to r. It is zero at the centre, where all directions balance, and it grows linearly as the Gaussian sphere encloses more charge. Put the two regions on one graph. Distance is measured in units of R, and field strength in units of its value at the surface. Inside, the field rises in a straight line from zero. In normalized form, E over E at the surface equals r over R. Enclosed charge grows as r cubed, while Gaussian area grows as r squared, leaving one power of r. At r equals R, the inside and outside formulas agree. There is no jump in the field because the charge fills a volume rather than sitting in an infinitesimally thin surface layer. Outside, E over the surface field equals R squared over r squared. The enclosed charge has stopped growing, while the Gaussian area continues to grow as r squared. The method was the same in both regions: use the source symmetry to choose a concentric sphere, make E constant on that surface, and then count only the charge actually enclosed."},{"title":"Lines, Sheets, and the Surfaces They Choose","startSeconds":586.4642083333332,"narration":"Now stretch the charge distribution into an ideal infinite line with uniform charge per length lambda. Translation along the line changes nothing, and rotation around it changes nothing. Those symmetries force the electric field to point directly away from the line. Its magnitude can depend on perpendicular distance r, but not on position along the line or angle around it. Choose a cylinder centred on the line. Its curved side is everywhere the same distance r from the charge, so E has one constant magnitude there. Let the solid settle into view. The charged line is the cylinder's axis, and the blue field arrows point through its curved wall. On the two end caps, the outward normals point along the line, while the electric field points radially away from it. Their dot product is zero, so the caps contribute no flux. Only the curved side contributes. Its area is circumference two pi r times length L, so the flux is E times two pi r L. The cylinder encloses a length L of line charge. Charge per length lambda times L gives enclosed charge lambda L. Apply Gauss's law. The length L appears on both sides and cancels. Solving leaves lambda over two pi epsilon zero r. The line field falls as one over r, not one over r squared. As the cylinder expands, its relevant area per unit length grows only in proportion to r. Next spread charge uniformly across an ideal infinite sheet, with surface charge density sigma. Sliding anywhere within the sheet cannot change the field. Rotating the sheet within its own plane also changes nothing. The only distinguished direction is perpendicular to the sheet, so the field must point normally away on both sides. Choose a short cylindrical pillbox that straddles the sheet. Its flat caps are parallel to the charge distribution, and its curved wall joins them. A small turn shows the construction. The field passes through the two caps, while it runs parallel to the curved wall. The curved wall contributes zero flux because its normal lies within the sheet while E is perpendicular to it. Each cap contributes E A, so the total flux is two E A. The pillbox encloses sheet area A, so it encloses charge sigma A. Gauss's law gives two E A equals sigma A over epsilon zero. The cap area cancels, leaving E equals sigma over two epsilon zero. There is no distance in the answer. For an ideal infinite sheet, moving the caps farther away does not spread a fixed bundle over a growing area. The same cap area intercepts the same flux. Put the three geometries together. A point spreads flux over a sphere whose area grows as r squared, so its field has the form a constant over r squared. A line spreads flux over the curved wall of a cylinder. Per unit length, that area grows as r, so the field has the form a constant over r. A sheet sends flux through two equal caps. Their area does not change when the pillbox grows taller, so the ideal sheet field is constant. Gauss's law was identical in all three cases. What changed was the symmetry, and symmetry determined the surface on which E became constant and the unwanted pieces contributed zero."},{"title":"Conductors, Surface Charge, and the Car","startSeconds":816.7534166666665,"narration":"A conductor contains mobile charge. If an electric field existed inside the metal, that charge would feel a force and begin to move. A state with moving charge is not electrostatic equilibrium. The mobile charge rearranges on an extremely short time scale. Its own electric field opposes the field that drove the motion, and equilibrium is reached only when the net electric field inside the conducting material is zero. That statement comes from the physics of mobile charge, not from Gauss's law alone. Gauss's law now tells us an important consequence of the zero field. Draw any closed Gaussian surface lying completely inside the conducting material. Because E is zero at every point on it, its electric flux is zero. Gauss's law then says that the net charge enclosed by every such interior surface is zero. Excess charge therefore cannot remain distributed through the bulk of the metal in electrostatic equilibrium. It moves outward and comes to rest on the conductor's surface. Outside the conductor, those surface charges can produce an electric field. Inside the conducting material, they have arranged themselves so that their combined field cancels. The surface distribution need not be uniform. Charge crowds more strongly near sharp points, where the surrounding field can become especially large. But the equilibrium condition inside the metal remains E equals zero. Now replace the simple conductor with the metal body of a hard-top car. The body forms a conducting shell around the passenger compartment. Suppose lightning strikes the roof. The lightning delivers charge and a large current to the exterior metal. The charge spreads over the outside, and the current finds conducting paths along the exterior body. The metal shell carries the dangerous electrical disturbance around the passenger space rather than through it. The passenger compartment is therefore close to one electric potential, with a strongly reduced electric field inside. This shielding behavior is often called the Faraday-cage effect. The protection comes from the continuous metal shell, not primarily from the rubber tires. During a storm, occupants should remain inside with windows closed and avoid touching metal parts connected to the exterior. The same principle is used deliberately in shielded rooms, cable coverings, and metal enclosures around sensitive electronics. Conductors rearrange charge so that their protected interiors experience very little electric field. Three ideas organize the whole lecture. First, electric flux is the signed amount of electric field crossing a surface, and the flux through a closed surface counts net enclosed charge. Second, Gauss's law becomes a field-solving tool only when symmetry chooses a useful surface: a sphere for spherical charge, a cylinder for a line, and a pillbox for a sheet. Third, mobile charge in a conductor rearranges until the field inside the metal is zero. Excess charge lives on the surface, and a closed metal body redirects an external electrical disturbance around its interior. Flux made the counting picture precise. Gauss's law connected that count to charge. Symmetry turned the law into three electric fields, and electrostatic equilibrium turned it into protection inside a conductor."}]}}