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Question: Answered & Verified by Expert
A spherical conducting shell of inner radius ' $r_1$ ' and outer radius ' $r_2$ ' has a charge ' $Q$ '. A charge $-q$ is placed at the center of the shell. The surface charge density on the inner and outer surface of the shell will be

PhysicsElectrostaticsJEE Main
Options:
  • A $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$
  • B $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}}{4 \pi \mathrm{r}_2^2}$
  • C $\frac{-\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}+\mathrm{q}}{4 \pi \mathrm{r}_2^2}$
  • D zero and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$
Solution:
2534 Upvotes Verified Answer
The correct answer is: $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$
Due to charge $-q$ at the center of the shell, a charge $q$ will be induced on the inner surface and $-q$ on the outer surface. The charge on outer surface will become $Q-q$. Hence surface charge densities will be $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$

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