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 in limits 589.0V 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 ?

Options

A. 589.4V
B. 589.5V
C. 289.2V
D. 289.6V

Answer

To solve the problem, we need to determine the electric potential at a point on a sphere whose radius vector makes an angle of 60 degrees with the direction of a uniform electric field.

### Step-by-step Solution:

1. **Understand the Potential Variation**:
The potential at various points on the sphere varies between 589.0 V and 589.8 V. This means the potential difference (ΔV) across the sphere is:
\[ \Delta V = 589.8 \, \text{V} - 589.0 \, \text{V} = 0.8 \, \text{V} \]

2. **Relate Potential Difference to Electric Field**:
The potential difference in a uniform electric field can be expressed as:
\[ \Delta V = -E \cdot d \]
where \(E\) is the magnitude of the electric field and \(d\) is the displacement in the direction of the electric field.

3. **Consider the Angle**:
Since we need the potential at a point where the radius vector makes an angle of 60 degrees with the direction of the electric field, we can express the relationship as:
\[ \Delta V = -E \cdot d \cdot \cos(\theta) \]
where \(\theta = 60^\circ\).

4. **Substituting Values**:
From the potential difference calculated, we can write:
\[ 0.8 = E \cdot d \cdot \cos(60^\circ) \]
Knowing that \(\cos(60^\circ) = \frac{1}{2}\), we can substitute this into the equation:
\[ 0.8 = E \cdot d \cdot \frac{1}{2} \]
Rearranging gives:
\[ E \cdot d = 0.8 \cdot 2 = 1.6 \]

5. **Calculate the Potential at the Point**:
The potential at the point making an angle of 60 degrees with the electric field can be calculated as:
\[ V = V_{\text{min}} + \Delta V \]
where \(V_{\text{min}} = 589.0 \, \text{V}\) (the minimum potential). The change in potential (\(\Delta V\)) at this point is:
\[ \Delta V = E \cdot d \cdot \cos(60^\circ) = 1.6 \cdot \frac{1}{2} = 0.8 \]
Therefore, the potential at this point is:
\[ V = 589.0 \, \text{V} + 0.4 \, \text{V} = 589.4 \, \text{V} \]

### Final Answer:
The potential at the point on the sphere whose radius vector makes an angle of 60 degrees with the direction of the field is **589.4 V**.