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. 589.5 V
C. 589.2 V
D. 589.6 V

Answer

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

### Step-by-Step Solution:

1. **Identify the given values**:
- The potential at various points on the sphere varies between 589.0 V and 589.8 V.
- The maximum potential difference (ΔV) is calculated as:
\[ ΔV = 589.8 \, \text{V} - 589.0 \, \text{V} = 0.8 \, \text{V} \]

2. **Determine the potential at the center of the sphere**:
- Since the potential varies, we can take the average potential at the center (point P) of the sphere:
\[ V_P = 589.0 \, \text{V} \quad (\text{assuming this is the potential at point P}) \]

3. **Calculate the potential difference due to the angle**:
- The potential difference between two points in an electric field can be expressed as:
\[ ΔV = -E \cdot d \cdot \cos(\theta) \]
- Here, \(E\) is the electric field strength, \(d\) is the distance, and \(\theta\) is the angle between the radius vector and the direction of the electric field. In this case, \(\theta = 60^\circ\).

4. **Find the effective potential drop**:
- The maximum potential drop occurs in the direction of the electric field, which is:
\[ \text{Maximum potential drop} = 0.8 \, \text{V} \]
- Since \(\cos(60^\circ) = \frac{1}{2}\), the potential drop at the angle of \(60^\circ\) is:
\[ ΔV_{60^\circ} = 0.8 \, \text{V} \cdot \cos(60^\circ) = 0.8 \, \text{V} \cdot \frac{1}{2} = 0.4 \, \text{V} \]

5. **Calculate the potential at the point making an angle of \(60^\circ\)**:
- The potential at the point making an angle of \(60^\circ\) with the electric field is:
\[ V_{60^\circ} = V_P - ΔV_{60^\circ} = 589.0 \, \text{V} - 0.4 \, \text{V} = 588.6 \, \text{V} \]

6. **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:
\[ V_{60^\circ} = 588.6 \, \text{V} \]