Question

In the given fig. the charge appears on the sphere is –

• A. q
• B. $\frac{qd}{r}$
• C. $–\frac{qr}{d}$
• D. Zero

Answer 1

Answer: (C)

$–\frac{qr}{d}$

Answer 2

The net potential on the surface of earthed conductor is zero.

\ V = $\frac{q_{1}}{4\pi\varepsilon_{0}r} + \frac{q}{4\pi\varepsilon_{0}d} = 0$

$\frac{q_{1}}{4\pi\varepsilon_{0}r} = –\frac{q}{4\pi\varepsilon_{0}d}$

$\Rightarrow$ q₁ = – $\frac{qr}{d}$