Physics · Ch 1 — Electrostatics
Applications of Gauss law
Applications of Gauss law
Although the electric field due to any given charge configuration can, in principle, always be computed either by direct application of Coulomb's law (using the superposition and integration techniques of sections 1.3.2-1.3.3) or by Gauss's law, Gauss's law is by far the more efficient method whenever the charge distribution has enough symmetry to allow a Gaussian surface to be chosen on which the field's magnitude is constant. Four standard cases are worked out. (i) An infinitely long straight wire with uniform linear charge density lambda: using a coaxial cylindrical Gaussian surface of radius r and length l, only the curved side contributes flux (E constant and radial there, zero flux through the flat ends), giving E(2 pi r l) = lambda l/epsilon0, so E = lambda/(2 pi epsilon0 r) -- the field falls off as 1/r. (ii) An infinite charged plane sheet with uniform surface charge density sigma: using a small pillbox Gaussian surface straddling the sheet, flux crosses only the two flat end caps of area A each, giving 2EA = sigma A/epsilon0, so E = sigma/(2 epsilon0) -- a field of constant magnitude, independent of distance from the sheet. (iii) Two parallel sheets carrying equal and opposite surface charge densities +sigma and -sigma: their individual uniform fields add in the region between the sheets (giving a combined field E = sigma/epsilon0 there) and exactly cancel everywhere outside the pair, so the net field is confined entirely to the gap between the sheets -- the idealised model used for a parallel-plate capacitor. (iv) A uniformly charged spherical shell of radius R carrying total charge Q: choosing a spherical Gaussian surface of radius r outside the shell (r > R) encloses the full charge Q and gives E = kQ/r^2, ex …
What this figure shows. A straight wire of infinite length, carrying a uniform linear charge density lambda (charge per unit length), is drawn with a field point P marked at perpendicular distance r from the wire, and a field vector E drawn pointing straight outward from the wire, perpendicular to it, at P. Because the wire's own symmetry guarantees the field can only point radially outward from the wire (never along the wire's length, by the mirror symmetry of an infinite wire) and can only depend on the perpendicular distance r (never on position along the wire's length), this figure sets up exactly the symmetry needed to choose a …
What this figure shows. A closed cylinder of radius r and length l is drawn coaxial with the infinite charged wire, so that the wire runs exactly along the cylinder's central axis; the field E is shown pointing radially outward and constant in magnitude everywhere on the cylinder's curved side surface (since every point on that curved surface is the same perpendicular distance r from the wire), while the field is shown running exactly parallel to (in the plane of) the cylinder's two flat end caps, contributing zero flux through them. This choice of Gaussian surface -- a coaxial cylinder -- is precisely what makes the Gauss's-law calcu …
Worked out. Choosing a cylindrical Gaussian surface of radius r and length l, coaxial with the infinite wire, the flux through the two flat end caps is zero (E runs parallel to them, never crossing), and the flux through the curved side surface is simply E multiplied by that surface's area, 2 pi r l, since E is constant in magnitude and everywhere perpendicular (radially outward) to the curved surface. The charge enclosed by this cylinder is lambda l (the linear charge density times the enclosed length). Gauss's law then gives E(2 pi r l) = lambda l/epsilon0, and the length l cancels from both sides, leaving E = lambda/(2 pi epsilon0 r) -- the field of an infinite charged wire falls off as 1/r (the first power of distance), which is a much slower falloff than the 1/r^2 of a single point charge, because the wire's char …
What this figure shows. An infinite flat sheet, carrying a uniform surface charge density sigma (charge per unit area), is drawn edge-on, with field vectors E drawn pointing straight away from the sheet on both sides, perpendicular to the plane, and of exactly equal magnitude on either side by the mirror symmetry of an infinite plane. A small pillbox-shaped Gaussian surface (a short cylinder) straddling the sheet, with its two flat faces parallel to the sheet, is superimposed to show the surface used to apply …
Worked out. A small cylindrical (pillbox) Gaussian surface is chosen straddling the sheet, with its two flat circular end caps, each of area A, parallel to the sheet and located symmetrically on either side of it. By symmetry, E is perpendicular to the sheet and has the same magnitude on both sides, so flux crosses only the two flat end caps (the curved side of the pillbox contributes nothing, since E runs parallel to it), giving total flux 2EA. The enclosed charge is sigma A. Gauss's law gives 2EA = sigma A/epsilon0, so E = sigma/(2 epsilon0), independent of the distance from the sheet -- the field of an infinite charged sheet is perfectly uniform, the same strength at every distance, unlike the field of a point charge or a line charge, both o …
What this figure shows. Two large parallel sheets are drawn facing each other a small distance apart, one carrying uniform surface charge density +sigma and the other -sigma; between the two sheets the individual fields from each sheet point the same way and add together, while outside the pair (above the positive sheet or below the negative sheet) the two individual fields point opposite ways and cancel exactly. The net result, shown by field-line arrows only in the gap between the sheets and none outside, is that the combined field is confined entirely to the region between the two sheets -- exactly the idealised configuration us …
What this figure shows. A hollow spherical shell of radius R, carrying charge Q spread uniformly over its surface, is drawn in panel (a); in panel (b), two concentric spherical Gaussian surfaces are superimposed, one of radius r less than R (entirely inside the shell) and one of radius r greater than R (outside the shell), each enclosing a different amount of charge -- the inner Gaussian sphere encloses zero net charge (all the actual charge lies out on the shell itself, outside this inner sphere), while the outer Gaussian sphere encloses the full charge Q. This is exactly the pair of cases needed to derive that the field is zero everywhere inside a uniformly charged shell and equals kQ/r^2, as if the whole charge were concentrated at the ce …
Worked out. For a spherical Gaussian surface of radius r drawn outside the shell (r greater than R), the enclosed charge is the shell's full charge Q, and by the shell's spherical symmetry E is constant in magnitude and radial everywhere on this Gaussian sphere, giving E(4 pi r^2) = Q/epsilon0, so E = kQ/r^2 -- identical to the field of a point charge Q located at the centre, even though the real charge is spread over the shell's surface, not concentrated at a point. For a spherical Gaussian surface drawn inside the shell (r less than R), the enclosed charge is exactly zero, since none of the shell's charge lies inside a sphere of radius smaller than R; Gauss's law then immediately gives E(4 pi r^2) = 0/epsilon0 = 0, so the field is exactly zero everywhere strictly inside a uniformly charged spherical shell, regardless of how close to the shell' …
What this figure shows. A graph with the field magnitude E on the vertical axis and the radial distance r on the horizontal axis shows the field staying flat at exactly zero for all r less than R (inside the shell), then jumping discontinuously up to its maximum value kQ/R^2 exactly at r = R (on the shell's surface), and finally falling off smoothly along a 1/r^2 curve for all r greater than R (outside the shell), asymptotically approaching zero as r goes to infinity. The graph is the single clearest visual summary of the two-part result derived by applying Gauss's law separat …