Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Open MARKS Web Download App
Search any question & find its solution
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

PhysicsElectrostaticsMHT CETMHT CET 2021 (20 Sep Shift 2)
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:
2220 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 $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.

Use on iOS or Desktop Download from Google Play