# Electric Field due to Uniformly Charged Thin Spherical Shell

## Problems from IIT JEE

**Problem (IIT JEE 2006):**
For spherical symmetrical charge distribution, variation of electric potential, $V$, with distance from centre, $r$, is given in figure. Which of the following option(s) is (are) correct,

- Total charge within $2R_0$ is $q$.
- Total electrostatic energy for $r\leq R_0$ is zero.
- At $r=R_0$ electric field is discontinuous.
- There will be no charge anywhere except at $r=R_0$.

**Solution:**
Use relation, $E=-\frac{\mathrm{d}V}{\mathrm{d}r}$, to get electric field for given potential,
\begin{align}
&V(r)=
\begin{cases}
\frac{q}{4\pi\epsilon_0 R_0} & \text{if $r\leq R_0$,} \\
\frac{q}{4\pi\epsilon_0 r} & \text{if $r>R_0$}, \\
\end{cases}\quad; &&
E(r)=
\begin{cases}
0 & \text{if $r\leq R_0$,} \\
\frac{q}{4\pi\epsilon_0 r^2} & \text{if $r>R_0$.} \\
\end{cases} \nonumber
\end{align}
Consider a spherical shell of radius $2R_0$ as Gaussian surface. Using Gauss's law for this surface, we get,
\begin{alignat}{2}
&q_\text{enc}=\epsilon_0\phi=\epsilon_0\cdot \frac{q}{16\pi\epsilon_0 R_0^2} \cdot 16\pi R_0^2=q.\nonumber
\end{alignat}
The electrostatic energy density for $r\leq R_0$ is $\frac{1}{2}\epsilon_0 E^2=0$ (since $E=0$) and hence total electric energy stored in this region is zero.
The electric field is discontinuous at $r=R_0$. This can be seen by taking left and right limits,
\begin{align}
&\lim_{r\to R_0^-}E(r)=0, &&\lim_{r\to R_0^+}E(r)=\frac{q}{4\pi \epsilon_0 R_0^2}. \nonumber
\end{align}
Since electric field is continuous at all points except at $r=R_0$, there is no charge distribution except at $r=R_0$. The readers are encouraged to realize that given potential is same as the potential due to a spherical shell of radius $R_0$ having a charge $q$.