A thin spherical shell of radius $$R$$ has charge $$Q$$ spread uniformly over its surface. Which of the following graphs most closely represents the electric field $$E(r)$$ produced by the shell in the range $$0 \le r < \infty$$, where $$r$$ is the distance from the centre of the shell?

For a thin spherical shell the total charge $$Q$$ resides on the surface. We have to find the electric field $$E(r)$$ at a distance $$r$$ from the centre for two regions :

Case 1: $$0 \le r \lt R$$ (inside the shell)
By Gauss’s law, choose a concentric Gaussian sphere of radius $$r\lt R$$. The enclosed charge is zero, therefore

$$\oint \mathbf{E}\cdot d\mathbf{S}=E(4\pi r^{2})=0\; \Longrightarrow\; E=0$$

Thus the field is identically zero for every interior point. On the graph the line from $$r=0$$ up to but not including $$r=R$$ lies on the horizontal axis.

Case 2: $$r \gt R$$ (outside the shell)
A concentric Gaussian sphere of radius $$r$$ now encloses the full charge $$Q$$. Applying Gauss’s law again,

$$E(4\pi r^{2})=\frac{Q}{\varepsilon_{0}}\;\Longrightarrow\;E(r)= \frac{1}{4\pi\varepsilon_{0}}\;\frac{Q}{r^{2}}$$

This is the same inverse-square variation as for a point charge located at the centre. Hence, for $$r\gt R$$ the graph is a rectangular hyperbola starting from $$r=R$$ and tending to zero as $$r\rightarrow\infty$$.

Value at the surface: Just inside the surface, $$E_{\text{in}}=0$$. Just outside, $$E_{\text{out}}=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Q}{R^{2}}$$. Because of this sudden jump, the graph has an open circle at $$r=R$$ on the zero line and a filled dot at $$r=R$$ on the value $$\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Q}{R^{2}}$$, then follows the $$1/r^{2}$$ curve.

The only option that shows
• $$E=0$$ for $$0\le r\lt R$$,
• a discontinuous rise at $$r=R$$, and
• an inverse-square decay for $$r\gt R$$
is Option A.

Option A which is: the curve with a flat zero segment inside the shell, a jump at $$r=R$$, and a $$1/r^{2}$$ fall thereafter.