There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $$P$$, in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60° with the direction of the field?

We have a region in which the electro-static field is uniform. For any point whose position vector from the centre $$P$$ is $$\vec r$$ and which makes an angle $$\theta$$ with the field direction, the potential is given by the scalar-product form of the relation

$$V \;=\;V_P \;+\;(\vec E\!\cdot\!\vec r)\;=\;V_P \;+\;E\,r\cos\theta$$

where $$V_P$$ is the potential at the centre, $$E$$ is the magnitude of the uniform field and $$r$$ is the radius of the sphere. (We have taken the sign convention such that the point lying along the field direction has the larger potential; any other consistent convention finally leads to the same numeric answer.)

The observed extreme values on the sphere are

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

These extremes correspond to $$\theta=0^\circ$$ and $$\theta=180^\circ$$ respectively, because $$\cos 0^\circ=+1$$ and $$\cos 180^\circ=-1$$. Substituting these two cases in the general expression:

For $$\theta = 0^\circ$$, $$V_{\max}=V_P+E r$$.

For $$\theta = 180^\circ$$, $$V_{\min}=V_P-E r$$.

Adding the two equations eliminates $$E r$$ and gives

$$2V_P = V_{\max}+V_{\min} \;=\;589.8 + 589.0 \;=\;1178.8,$$

so

$$V_P = \dfrac{1178.8}{2}=589.4\;{\rm V}.$$

Subtracting the two extreme-value equations eliminates $$V_P$$ and yields

$$2 E r = V_{\max}-V_{\min} \;=\;589.8 - 589.0 \;=\;0.8,$$

hence

$$E r = 0.4\;{\rm V}.$$

Now we want the potential at a point for which $$\theta = 60^\circ$$. Using $$\cos 60^\circ = \dfrac12$$ in the basic formula, we get

$$V = V_P + E r \cos 60^\circ = 589.4 + (0.4)\!\left(\dfrac12\right) = 589.4 + 0.2 = 589.6\;{\rm V}.$$

Hence, the correct answer is Option D.