Physics · Ch 1 — Electrostatics
Gauss law
Gauss law
A positive point charge Q is imagined surrounded by a sphere of radius r centred exactly on it. Because every point of the sphere's surface is the same distance r from Q, the field E has the same constant magnitude k Q/r^2 at every point on the sphere, and, since the field of a point charge is always radial, E is everywhere exactly parallel to the sphere's own outward normal. This special symmetry makes the flux integral trivial: Phi_E = the surface integral of E . dA = E multiplied by the sphere's total surface area (since E is constant and parallel to dA everywhere) = (k Q/r^2)(4 pi r^2) = Q/epsilon0 (using k = 1/4 pi epsilon0). Remarkably, the r^2 in the field's denominator exactly cancels the r^2 in the sphere's surface area, so the result Phi_E = Q/epsilon0 does not depend on the radius r of the chosen sphere at all. A further geometric argument -- every field line leaving the point charge Q must eventually cross any closed surface that fully encloses it, however irregularly shaped that surface is, since field lines can only terminate on a charge -- extends this result from the special case of a sphere to a closed surface of completely arbitrary shape, and by superposition, to any number and arrangement of enclosed point charges (and, by extension, continuous charge distributions). This is Gauss's law: for any closed surface, the total (net) electric flux through it equals the total (net) charge enclosed within it, divided by epsilon0: Phi_E = Q_enclosed / epsilon0. Charges located outside the chosen closed surface contribute zero net flux (their field lines pass into the surface on one side and back out on the other, cancelling exactly), so only the enclosed charge …
What this figure shows. A single positive point charge Q sits at the centre of an imaginary sphere of radius r; because every point on the sphere's surface is the same distance r from the charge, the field E has exactly the same magnitude everywhere on the sphere, and it points radially outward everywhere, exactly parallel to the sphere's own outward normal at every point. This is precisely the special symmetry that makes the flux integral trivial to evaluate -- since E and dA are parallel everywhere and E is constant in magnitude over the whole surface, the flux integral collapses to simply E multiplied by the sphere's total surface area, 4 pi r^2, giving Phi_E = E(4 pi r^2) = Q/epsilon0 af …
What this figure shows. The same point charge Q is now surrounded by a second, irregularly shaped closed surface (not a sphere) drawn around it, together with the original sphere shown enclosed inside it for comparison; the same field lines that pass outward through the sphere are shown continuing on to pass through the irregular surface as well, since every field line that leaves the charge must eventually cross any surface that fully encloses it, however oddly shaped that surface is. This visual argument is the basis for extending the flux result Phi_E = Q/epsilon0, proved rigorously only for the easy, symmetric case of a sphere, to the fully general statement of Gauss's law that holds for a c …