Question

The adjacent figure shows variation of electric potential with distance for a spherical symmetric charge distribution system and given as $\phi_r = \frac{q}{4\pi\epsilon_0 r}(r \geq R_0)$, $\phi_r = \frac{q}{4\pi\epsilon_0 R_0}(r \leq R_0)$. Which of the following option is/are correct -

• A. For spherical region $r \leq R_0$ total electrostatic energy stored is zero
• B. Within $r = 2R_0$, total charge is q
• C. There will be no charge anywhere except at $r = R_0$
• D. Electric fields disontinuous at $r = R_0$

Answer 1

Answer: (C)

There will be no charge anywhere except at $r = R_0$

Answer 2

The electric potential is constant for $r \leq R_0$ and decreases as $1/r$ for $r \geq R_0$. This potential distribution is characteristic of a charged conducting sphere or a spherical shell with charge $q$ on its surface at $r=R_0$. Inside a conductor, the electric field is zero, and the potential is constant. The charge in a conductor resides on its surface. Outside a spherically symmetric charge distribution, the potential is the same as that of a point charge at the center if the total charge is $q$.

From the given potential, the electric field for $r < R_0$ is $E_r = -\frac{d}{dr}(\text{constant}) = 0$. The electric field for $r > R_0$ is $E_r = -\frac{d}{dr}(\frac{q}{4\pi\epsilon_0 r}) = \frac{q}{4\pi\epsilon_0 r^2}$.

Since the electric field is zero for $r < R_0$, by Gauss's law, there is no net charge within any sphere of radius $r < R_0$.

Since the electric field for $r > R_0$ is the same as that of a point charge $q$ at the origin, the total charge enclosed within any sphere of radius $r > R_0$ is $q$.

Since there is no volume charge density for $r < R_0$ and for $r > R_0$, and the total charge is $q$, the entire charge must be located at $r = R_0$ as a surface charge. Therefore, there will be no charge anywhere except at $r = R_0$.