A hollow spherical conductor of radius r potential of 100 V at its outer surface. The potential inside the hollow at a distance of` (r )/(2)` from its centre is

Options

A. 100 V
B. 50 V
C. 200 V
D. zero

Answer

To solve the problem, we need to determine the electric potential inside a hollow spherical conductor at a distance of \( \frac{r}{2} \) from its center, given that the potential at the outer surface (radius \( r \)) is 100 V.

### Step-by-Step Solution:

1. **Understanding the Properties of a Hollow Spherical Conductor**:
- A hollow spherical conductor has the property that the electric field inside the hollow region is zero. This is due to the symmetry of the charge distribution on the surface of the conductor.

2. **Electric Field Inside the Conductor**:
- Since the electric field \( E \) inside the hollow region is zero, we can express this mathematically:
\[ E = -\frac{dV}{dr} = 0 \]
- This implies that the potential \( V \) does not change with respect to the distance \( r \) inside the hollow region.

3. **Constant Potential Inside the Hollow Sphere**:
- Because the electric field is zero, the potential inside the hollow conductor remains constant throughout the entire hollow region. This means that the potential at any point inside the hollow sphere is equal to the potential at the surface.

4. **Given Potential at the Surface**:
- We are given that the potential at the outer surface (at radius \( r \)) is 100 V. Therefore, the potential at any point inside the hollow sphere, including at a distance of \( \frac{r}{2} \) from the center, is also 100 V.

5. **Conclusion**:
- Hence, the potential inside the hollow at a distance of \( \frac{r}{2} \) from its center is:
\[ V = 100 \, \text{V} \]

### Final Answer:
The potential inside the hollow at a distance of \( \frac{r}{2} \) from its center is **100 V**.

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