Physics · Ch 10 — Electrostatics

Gauss' Law

10.8

Gauss' Law

Karl Friedrich Gauss (1777-1855), one of the greatest mathematicians of all time, formulated a law -- now called GAUSS' LAW -- expressing a deep, general relationship between electric charge and the electric field it produces. Gauss' law is, in a sense, analogous to Coulomb's law: both connect electric field to electric charge. But where Coulomb's law gives the field due to a SINGLE point charge directly, Gauss' law provides an equivalent, often far more convenient, alternative method for finding the field: it relates the total field summed (integrated) over an entire CLOSED SURFACE to the total charge enclosed WITHIN that surface.

Derivation. Consider a closed surface of any arbitrary shape, enclosing some number of positive electric charges (Fig. 10.15). To derive Gauss' theorem, imagine first a single small charge $+q$ present at some interior point O, and imagine a small, infinitesimal area element dA somewhere on the given (possibly irregular) closed surface. The magnitude of the electric field at the point P on this element dA, due to the charge $+q$ at O, is $E=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r^2}$ (r being the distance OP), directed radially away from O. Let $\theta$ be the angle between the OUTWARD NORMAL to the area element dA and the direction of E at that point; the electric flux $d\phi$ passing through this element is $d\phi=E\cos\theta\,dA=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta\,dA=\frac{q}{4\pi\epsilon_0}\,d\Omega,$ where $d\Omega=\dfrac{dA\cos\theta}{r^2}$ is the SOLID ANGLE that the small area dA subtends at the point O.

The TOTAL electric flux $\phi_E$ through the entire closed surface is found by integrating this expression over the whole surface: $\phi_E=\oint_S d\phi=\oint\vec{E}\cdot d\vec{S}=\frac{q}{4\pi\epsilon_0}\oint d\Omega=\frac{q}{4\pi\epsilon_0}(4\pi)=\frac{q}{\epsilon_0},$ since the total solid angle subtended by ANY complete closed surface at an interior point is always exactly $4\pi$ steradians, regardless of the surface's actual shape. This is Gauss' theorem for a single enclosed charge, and it holds true for EVERY individual electric charge enclosed inside a given closed surface -- so if separate charges $q_1,q_2,\dots,q_n$ are all enclosed inside the same surface, the flux due to each ALONE is $q_1/\epsilon_0,q_2/\epsilon_0,\dots,q_n/\epsilon_0$ respectively.

A positive sign in this result indicates the flux is directed OUTWARD, away from the enclosed positive charge (Fig. 10.16a); if the enclosed charge is negative, the flux is directed INWARD instead (Fig. 10.16b); and if a charge sits OUTSIDE the closed surface entirely, the NET flux due to it through that surface works out to exactly zero, since every field line from that outside charge that enters the surface on one side must also leave it again on another (Fig. 10.16c).

By the principle of superposition, the total flux due to ALL charges enclosed within a given closed surface is simply the algebraic SUM of the individual contributions: $\phi_E=\frac{q_1}{\epsilon_0}+\frac{q_2}{\epsilon_0}+\dots+\frac{q_n}{\epsilon_0}=\frac{1}{\epsilon_0}\sum_{i=1}^{n}q_i=\frac{Q}{\epsilon_0}.$ …

Figure 10.15Fig. 10.15: Setting up Gauss' law -- a point charge inside an irregular closed surface

What this figure shows. An irregularly-shaped closed surface (drawn as a lumpy, non-spherical blob) encloses a small positive point charge +q marked at an interior point O. An infinitesimal patch of area dA is marked on the irregular surface at a point P, with the electric field E at P (due to +q) drawn as an arrow pointing radially outward from O through P, and the angle $\theta$ between this field direction and the local normal to the small area dA is indicated -- this is the exact geometric setup used to derive $d\phi=\frac{q}{4\pi\epsilon_0}\,d\Omega$ (with $d\Omega$ the solid angle subtended by dA at O), before integrating over the whole irregular surface to get the general resu …

Figure 10.16Fig. 10.16 (a)-(c): Flux due to a positive charge, a negative charge, and a charge outside the surface

What this figure shows. Three panels, each showing one closed surface (e.g. drawn as a sphere or irregular blob) with field lines crossing it: (a) a POSITIVE charge enclosed inside the surface, with field lines shown passing OUTWARD through the surface at every crossing point, illustrating positive (outward) flux, $+q/\epsilon_0$ ; (b) a NEGATIVE charge enclosed inside the surface, with field lines shown passing INWARD through the surface, illustrating negative (inward) flux; (c) a charge placed OUTSIDE the closed surface entirely, with field lines shown entering the surface on one side and leaving again on the other side (each line that enters also exits), so that the NET flux through the whole closed surface is exactly zero, even though individual fi …

Misc Ex.8Example 10.8: Flux through a sphere enclosing a point charge, and its independence from the sphere's radius

Worked out. A charge of $q=5.0$ C sits at the centre of a sphere of radius 1 m. By Gauss' law, the total flux through the sphere is $\phi=q/\epsilon_0=5.0/(8.85\times10^{-12})\approx5.65\times10^{11}$ Vm (this can equivalently be found via $\phi=E\times4\pi r^2$ , since E is constant over the sphere and everywhere parallel to the outward area vector). If the sphere's radius is instead doubled to 2 m, the flux is UNCHANGED, still $\approx5.65\times10^{11}$ Vm, because $E\propto1/r^2$ while the sphere's area $\propto r^2$ , so their product (and hence the flux) stays exactly constant -- exactly as Gauss' law predicts, since the enclosed charge itself has not changed and the law shows flux depends on enclosed charge alone, never on the size or shape of the …

Figure 10.17Fig. 10.17: Flux is independent of the shape and size of the enclosing surface

What this figure shows. A single point charge at the centre, surrounded by two different closed surfaces drawn one around the other -- an inner sphere of one radius and an outer sphere of a larger radius (or, in a more general version, an inner sphere and an outer surface of some irregular shape) -- with the same set of field lines from the central charge shown passing through BOTH surfaces. The figure visually confirms that the identical number of field lines (hence the identical total flux) crosses both the smaller and the larger enclosing surface, reinforcing that total flux depends only on the enclosed charge, never on how big or what shape the su …

Misc DYK.3Do you know? -- The Gaussian surface

Worked out. Explains that although the total flux through ANY closed surface enclosing a charge is the same (always $q/\epsilon_0$ , by Gauss' law), evaluating the flux INTEGRAL $\oint\vec{E}\cdot d\vec{S}$ directly is only easy when a surface is chosen to match the natural SYMMETRY of the charge distribution -- such a conveniently symmetric, imaginary, purely mathematical closed surface is called a Gaussian surface. For a single point charge the natural Gaussian surface is a sphere centred on the charge (so E is constant in magnitude and always parallel to the surface's normal everywhere on it); for a uniformly charged infinite line, it is a coaxial cylinder. A Gaussian surface has no physical existen …

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