A nonuniformly charged ring is kept near an uncharged conducting solid sphere. The distance between their centres (which are on the same line normal to the plane of the ring) is 3m and their radius is 4m. If total charge on the ring is `1muC`, then the potential of the sphere will be

Options

A. `2.25 kV`
B. `3 kV`
C. `1.8 kV`
D. cannot be determined

Explanation

To solve the problem, we need to find the potential of an uncharged conducting solid sphere when placed near a nonuniformly charged ring. Here are the steps to arrive at the solution:

### Step 1: Understand the Configuration
We have a nonuniformly charged ring with a total charge \( Q = 1 \mu C = 1 \times 10^{-6} C \) and a conducting solid sphere that is uncharged. The distance between the centers of the ring and the sphere is \( d = 3 m \), and the radius of the ring is \( R = 4 m \).

### Step 2: Determine the Distance from the Ring to the Sphere
Since the distance between the centers is given as \( 3 m \) and the radius of the ring is \( 4 m \), we can visualize that the closest point of the ring to the sphere is at a distance of \( d - R = 3 m - 4 m = -1 m \). However, this indicates that the ring is actually further away from the sphere than its radius, so we will consider the distance from the center of the ring to the surface of the sphere.

### Step 3: Calculate the Effective Distance
The effective distance from the ring to the surface of the sphere is:
\[ D = d + R = 3 m + 4 m = 7 m \]

### Step 4: Use the Formula for Electric Potential
The potential \( V \) at a point due to a point charge is given by:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r} \]
where \( r \) is the distance from the charge to the point where the potential is being calculated, and \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \).

### Step 5: Substitute Values into the Formula
Substituting \( Q = 1 \times 10^{-6} C \) and \( r = 7 m \):
\[ V = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{1 \times 10^{-6}}{7} \]

### Step 6: Calculate the Numerical Value
Calculating \( \frac{1}{4 \pi (8.85 \times 10^{-12})} \):
\[ \frac{1}{4 \pi (8.85 \times 10^{-12})} \approx 9 \times 10^9 \, N \cdot m^2/C^2 \]
Now substituting this into the potential formula:
\[ V \approx 9 \times 10^9 \cdot \frac{1 \times 10^{-6}}{7} \approx \frac{9 \times 10^3}{7} \approx 1285.71 \, V \]

### Step 7: Convert to Kilovolts
To express the potential in kilovolts:
\[ V \approx 1.28571 \, kV \approx 1.29 \, kV \]

### Conclusion
The potential of the sphere due to the charged ring is approximately **1.29 kV**.