To solve the problem, we need to determine the electric potential at a point on a sphere where the radius vector makes an angle of \(60^\circ\) with the direction of a uniform electric field. The potential varies between 589.0 V and 589.8 V across the sphere.
### Step-by-Step Solution:
1. **Identify the Potential Range**:
The potential at various points on the sphere varies from 589.0 V to 589.8 V. This means that the minimum potential \(V_{\text{min}} = 589.0 \, \text{V}\) and the maximum potential \(V_{\text{max}} = 589.8 \, \text{V}\).
2. **Calculate the Change in Potential**:
The change in potential across the sphere can be calculated as:
\[
\Delta V = V_{\text{max}} - V_{\text{min}} = 589.8 \, \text{V} - 589.0 \, \text{V} = 0.8 \, \text{V}
\]
3. **Understanding the Electric Field**:
In a uniform electric field, the potential difference is related to the electric field \(E\) and the distance \(d\) moved in the direction of the field:
\[
\Delta V = -E \cdot d
\]
Here, \(d\) is the distance along the direction of the electric field.
4. **Using the Dot Product**:
Since the radius vector makes an angle of \(60^\circ\) with the electric field, we can express the potential change as:
\[
\Delta V = -E \cdot d \cdot \cos(60^\circ)
\]
Given that \(\cos(60^\circ) = \frac{1}{2}\), we can rewrite the equation as:
\[
\Delta V = -E \cdot d \cdot \frac{1}{2}
\]
5. **Relating Change in Potential to Electric Field**:
From the previous steps, we can set up the equation:
\[
0.8 = E \cdot d \cdot \frac{1}{2}
\]
Rearranging gives:
\[
E \cdot d = 1.6 \, \text{V}
\]
6. **Determine the Potential at the Point**:
Since the potential is maximum at the point closest to the direction of the electric field (which is at \(589.8 \, \text{V}\)), and we are moving away from this point at an angle of \(60^\circ\), we need to find the potential at this point:
\[
V = V_{\text{max}} - \Delta V
\]
Thus, substituting the values:
\[
V = 589.8 \, \text{V} - 0.4 \, \text{V} = 589.4 \, \text{V}
\]
### Final Answer:
The potential at the point on the sphere whose radius vector makes an angle of \(60^\circ\) with the direction of the electric field is \(589.4 \, \text{V}\).
