Electric Potential & Equipotentials question from JEE Main 2017 (8 Apr, Shift 1)

Answer: (c)

1. In a uniform field $V = V_P - Er\cos\theta$ on the sphere, $\theta$ measured from the field direction.
2. Max $589.8$ at $\theta = 180^\circ$, min $589.0$ at $\theta = 0$: $V_P = 589.4\,\text{V}$, $Er = 0.4\,\text{V}$.
3. At $\theta = 60^\circ$: $V = 589.4 - 0.4\cos 60^\circ = 589.4 - 0.2 = 589.2\,\text{V}$
4. So the working gives (b) 589.2 V.