A spherical metal shell A of radius \( r_{1} \) and a solid metal sphere B of radius \( r_{2} \) are kept far apart and each is given charge \( +Q \). When connected with a thin wire, then
(A) inside sphere A, electric field, \( E_{A} = 0 \)
(B) ratio of their surface charge densities \( \displaystyle\frac{\sigma_{A}}{\sigma_{B}} = \frac{r_{2}}{r_{1}} \)
(C) inside sphere B, electric field, \( E_{B} = 0 \)
(D) capacitance of both the spheres is same.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
When the two conductors are joined by a wire, charge flows until they are at the same potential. For an isolated sphere of radius \( r \), \( V = q/(4\pi\varepsilon_{0}r) \), so after joining:
\( \displaystyle \frac{q_{A}}{r_{1}} = \frac{q_{B}}{r_{2}} \)
(A) inside sphere A, electric field, \( E_{A} = 0 \) — Correct. A is a hollow metal shell. All its charge sits on the outer surface and there is no charge in the cavity, so the field inside a conductor (and inside its cavity) is zero.
(B) ratio of their surface charge densities \( \displaystyle\frac{\sigma_{A}}{\sigma_{B}} = \frac{r_{2}}{r_{1}} \) — Correct. With \( q \propto r \) and \( \sigma = q/(4\pi r^{2}) \), the surface charge density goes as \( 1/r \):
\( \displaystyle \frac{\sigma_{A}}{\sigma_{B}} = \frac{q_{A}/r_{1}^{2}}{q_{B}/r_{2}^{2}} = \frac{r_{1}}{r_{2}}\times\frac{r_{2}^{2}}{r_{1}^{2}} = \frac{r_{2}}{r_{1}} \)
(C) inside sphere B, electric field, \( E_{B} = 0 \) — Correct. B is a solid conductor in equilibrium, so the field inside it is zero.
(D) capacitance of both the spheres is same. — Incorrect. The capacitance of an isolated sphere is \( 4\pi\varepsilon_{0}r \). It depends only on the radius, so the two are equal only if \( r_{1} = r_{2} \), which is not given.