A hollow charged metal sphere has radius `r`. If the potential difference between its surface and a point at a distance `3r` from the centre is V, then electric field intensity at a distance `3r` is

Options

A. `(V)/(2r)`
B. `(V)/(3r)`
C. `(V)/(6r)`
D. `(V)/(4r)`

Answer

To solve the problem, we need to find the electric field intensity at a distance \(3r\) from the center of a hollow charged metal sphere with radius \(r\). We know that the potential difference between the surface of the sphere and the point at distance \(3r\) is \(V\).

### Step-by-Step Solution:

1. **Identify the Potential at the Surface and at Point \(P\)**:
- The potential \(V_s\) at the surface of the sphere (radius \(r\)) is given by:
\[ V_s = \frac{KQ}{r} \]
- The potential \(V_p\) at point \(P\) (distance \(3r\) from the center) is given by:
\[ V_p = \frac{KQ}{3r} \]

2. **Write the Expression for the Potential Difference**:
- The potential difference \(V\) between the surface and point \(P\) is:
\[ V = V_s - V_p = \frac{KQ}{r} - \frac{KQ}{3r} \]

3. **Simplify the Potential Difference**:
- Combine the terms:
\[ V = \frac{KQ}{r} - \frac{KQ}{3r} = \frac{3KQ}{3r} - \frac{KQ}{3r} = \frac{2KQ}{3r} \]

4. **Relate \(KQ/r\) to the Potential Difference \(V\)**:
- From the equation above, we can express \(KQ/r\) in terms of \(V\):
\[ \frac{2KQ}{3r} = V \implies KQ = \frac{3Vr}{2} \]

5. **Determine the Electric Field at Point \(P\)**:
- The electric field \(E\) at a distance \(3r\) from the center is given by:
\[ E = \frac{KQ}{(3r)^2} = \frac{KQ}{9r^2} \]

6. **Substitute \(KQ\) into the Electric Field Expression**:
- Substitute \(KQ = \frac{3Vr}{2}\) into the electric field equation:
\[ E = \frac{\frac{3Vr}{2}}{9r^2} = \frac{3V}{2 \cdot 9r} = \frac{3V}{18r} = \frac{V}{6r} \]

7. **Final Result**:
- Therefore, the electric field intensity at a distance \(3r\) from the center of the sphere is:
\[ E = \frac{V}{6r} \]

### Conclusion:
The electric field intensity at a distance \(3r\) from the center of the hollow charged metal sphere is \(\frac{V}{6r}\).