Question

Concentric, thin metallic spheres of radii $r_{1}$ and $r_{2}\left(r_{1} > r_{2}\right)$ carry charges $q_{1}$ and $q_{2}$, respectively . Then the electric potential at a distance $r\left(r_{2} < r < r_{1}\right)$ will be $\frac{1}{4 \pi \varepsilon_{0}}$ times

• A. $\frac{q_{0} q_{1}+q_{2}}{r}$
• B. $\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r}$
• C. $\frac{q_{1}}{r_{2}}+\frac{q_{2}}{r_{2}}$
• D. $\frac{q_{1}}{r_{2}}+\frac{q_{2}}{r_{1}}$

Answer 1

Answer: (B)

The given point is inside the larger sphere.
So, potential at this point is the same as on the surface of the sphere.
The value is $\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1}}$.
The given point is outside the smaller sphere.
So, the charge on the smaller sphere would behave as if concentrated at the centre. The potential due to smaller sphere is $\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r}$.
Applying principle of superposition of potentials, the total potential is $\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r}\right]$.