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 $\frac{1}{4\pi {\epsilon }_0}$ 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 $\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r_1}$.
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 $\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{r}$.
Applying principle of superposition of potentials, the total potential is $\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r}\right]$.