MHT CET · Physics · Electrostatics

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

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 \(\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}\) and \(\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

Step-by-step Solution

Detailed explanation

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 \(\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app