1. Potential variation on a small sphere in a uniform electric field

In a uniform electric field, potential changes linearly with displacement along the field. If the center of the sphere is at point $P$ and the sphere has radius $r$, then for a point on the sphere making angle $\theta$ with the direction of the field,

$V(\theta)=V_P-Er\cos\theta$

where $E$ is the magnitude of the electric field.

2. Use the given maximum and minimum potentials

The potential on the sphere varies from

$V_{\min}=589.0\,\text{V}, \qquad V_{\max}=589.8\,\text{V}$

For a sphere in uniform field:

• maximum potential occurs at $\theta=180^\circ$
• minimum potential occurs at $\theta=0^\circ$

So,

$V_{\max}=V_P+Er$ $V_{\min}=V_P-Er$

Adding,

$V_{\max}+V_{\min}=2V_P$

$V_P=\frac{589.8+589.0}{2}=589.4\,\text{V}$

Also,

$2Er=589.8-589.0=0.8$

$Er=0.4\,\text{V}$

3. Find the potential at $\theta=60^\circ$

Given that the radius vector makes angle $60^\circ$ with the direction of the field,

$V=V_P-Er\cos 60^\circ$

Since $\cos 60^\circ=\frac12$,

$V=589.4-0.4\times \frac12$

$V=589.4-0.2=589.2\,\text{V}$

4. Match with the options

$\boxed{589.2\,\text{V}}$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer is C: 589.4 V, but the derived answer is B: 589.2 V.

The stored answer appears incorrect because $589.4\,\text{V}$ is the potential at the center of the sphere, not at the point where the radius vector makes $60^\circ$ with the field direction.