Physics · Ch 1 — Electric Charges and Fields

Application of Gauss's Theorem: Field Due to a Uniformly Charged Thin Spherical Shell

1.17

Application of Gauss's Theorem: Field Due to a Uniformly Charged Thin Spherical Shell

The problem. Find the electric field due to a thin spherical shell of radius $R$ carrying a total charge $Q$ spread uniformly over its surface, both outside the shell and inside it.

Exploiting the symmetry. A uniformly charged spherical shell looks identical when viewed from any direction, so the field it produces must be purely radial (pointing either directly toward or directly away from the centre) and must have a magnitude that depends only on the distance $r$ from the centre, not on direction. This spherical symmetry is matched by choosing, as the Gaussian surface, an imaginary sphere of radius $r$ concentric with the shell -- and, because the answer will differ depending on whether $r$ is larger or smaller than the shell's own radius $R$ , the two cases must be treated separately.

Case 1: outside the shell ( $r > R$ ). Choose a Gaussian sphere of radius $r$ , lying entirely outside the charged shell. Since $\vec{E}$ is radial and constant in magnitude everywhere on this sphere (by symmetry), it is exactly parallel to the outward normal at every point, so the flux is simply $E$ times the sphere's total surface area:

\Phi = E\times4\pi r^2

The Gaussian sphere of radius $r>R$ encloses the entire shell, so $q_{\text{enc}}=Q$ , the full charge on the shell. Gauss's theorem gives

E(4\pi r^2) = \frac{Q}{\epsilon_0} \quad\Longrightarrow\quad \boxed{E = \frac{Q}{4\pi\epsilon_0 r^2} = \frac{kQ}{r^2}} \qquad (r>R)

This is exactly the same formula as for a single point charge $Q$ placed at the centre -- a uniformly charged shell behaves, from any point outside it, precisely as though all its charge were concentrated at a single point at its centre.

Case 2: inside the shell ( $r < R$ ). Choose a smaller Gaussian sphere of radius $r$ , lying entirely inside the shell. All of the shell's charge sits on its surface, at radius $R$ , which is outside this smaller Gaussian sphere -- so the Gaussian sphere of radius $r<R$ encloses no charge at all: $q_{\text{enc}}=0$ . Gauss's theorem then gives

E(4\pi r^2) = \frac{0}{\epsilon_0} = 0 \quad\Longrightarrow\quad \boxed{E = 0} \qquad (r<R)

The electric field is exactly zero everywhere inside a uniformly charged thin spherical shell, no matter how close to the surface the point considered might be (so long as it remains strictly inside). …

Figure 1Spherical Gaussian surfaces outside and inside a charged shell

What this figure shows. A thin spherical shell of radius $R$ is drawn as a circle (representing a sphere in cross-section), its boundary marked with evenly spaced plus signs to show a uniform positive surface charge density spread only on the shell itself, with the interior of the shell left completely blank (no charge inside). Two concentric dashed circles are drawn around the same centre as the shell: a larger dashed circle of radius $r_1 > R$ lying entirely OUTSIDE the shell, representing the Gaussian surface used for an exterior point, with short outward radial arrows labelled $E$ drawn at several points on this dashed circle, all of equal length; and a smaller dashed circle of radius $r_2 < R$ lying entirely INSIDE the shell, representing the Gaussian surface used for an interior point, drawn with no field arrows on it at all and a small label readin …

Table 1Summary: electric field formulas from Gauss's theorem

Configuration	Electric field $E$	Variation with distance $r$
Point charge $q$	$E = kq/r^2$	$\propto 1/r^2$
Infinite line charge (density $\lambda$ )	$E = \lambda/(2\pi\epsilon_0 r) = 2k\lambda/r$	$\propto 1/r$
Infinite plane sheet (density $\sigma$ )	$E = \sigma/(2\epsilon_0)$	constant, independent of $r$

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