Question 5

**Setting up:** q = 10⁻⁸ C, d = 20 cm, R = 10 cm, k = 9×10⁹. **(A)** Neutral isolated sphere: its potential equals the potential the point charge produces at its center (induced charges sum to zero and contribute nothing at the center). V = kq/d = (9×10⁹)(10⁻⁸)/0.2 = 450 V. ✔ **(B)** On grounding, V = 0: kQ/R + kq/d = 0 ⇒ Q = −qR/d = −(10⁻⁸)(0.1)/(0.2) = −5×10⁻⁹ C. The sphere acquires −5×10⁻⁹ C, i.e. a positive charge of 5×10⁻⁹ C effectively flows from the sphere to the ground. ✔ **(C)** After removing the ground, the charge −5×10⁻⁹ C remains trapped on the sphere. ✔ **(D)** Point charge moved to d' = 30 cm. Potential of the sphere: V = kQ/R + kq/d' = (9×10⁹)(−5×10⁻⁹)/0.1 + (9×10⁹)(10⁻⁸)/0.3 = −450 + 300 = −150 V. Not 300 V. ✘ **Correct options: A, B, C**

No written reasoning was returned for this item.