Question 5

Let $k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$ Nm$^2$/C$^2$, $q = 10^{-8}$ C, $R = 0.1$ m and initial distance $r_1 = 0.2$ m.

(A) Before grounding, the potential of the neutral conducting sphere is uniform and equal to the potential at its center due to the external charge $q$: $V = \frac{kq}{r_1} = \frac{9 \times 10^9 \times 10^{-8}}{0.2} = \frac{90}{0.2} = 450 \text{ V}$ So, statement (A) is correct.

(B) & (C) When the sphere is grounded, its potential becomes zero. Let $q'$ be the charge induced on the sphere: $V_{center} = \frac{kq}{r_1} + \frac{kq'}{R} = 0$ $450 + \frac{9 \times 10^9 \times q'}{0.1} = 0 \Rightarrow 450 + 9 \times 10^{10} q' = 0$ $q' = -\frac{450}{9 \times 10^{10}} = -5 \times 10^{-9} \text{ C}$ Since the sphere was initially neutral, the charge that flowed from the sphere to the ground is $q_{initial} - q' = 0 - (-5 \times 10^{-9}) = 5 \times 10^{-9}$ C. So, statements (B) and (C) are correct.

(D) After removing the ground, the charge $q'$ remains on the sphere. The external charge $q$ is moved to $r_2 = 20 + 10 = 30$ cm = $0.3$ m. The final potential of the sphere is: $V_{final} = \frac{kq}{r_2} + \frac{kq'}{R} = \frac{9 \times 10^9 \times 10^{-8}}{0.3} + \frac{9 \times 10^9 \times (-5 \times 10^{-9})}{0.1}$ $V_{final} = 300 - 450 = -150 \text{ V}$ So, statement (D) is incorrect.