A conducting sphere of radius `R` is given a charge `Q`. The electric potential and the electric field at the centre of the sphere respectively are

Options

A. zero and `Q/(4pi epsi_(0) R^(2))`
B. `Q/(4pi epsi_(0)R)` and zero
C. `Q/(4pi epsi_(0) R)` and `Q/(4pi epsi_(0)R^(2))`
D. Both are zero

Answer

To solve the problem, we need to determine the electric potential and electric field at the center of a conducting sphere with radius \( R \) and charge \( Q \).

### Step-by-Step Solution:

1. **Understanding the Properties of Conducting Spheres**:
- A conducting sphere distributes its charge uniformly on its surface.
- Inside a conducting sphere, the electric field is zero. This is due to the fact that the charges on the surface rearrange themselves in such a way that they cancel any electric field within the conductor.

2. **Electric Field at the Center**:
- Since the electric field inside a conductor is zero, the electric field at the center of the sphere is:
\[ E = 0 \, \text{N/C} \]

3. **Electric Potential at the Center**:
- The electric potential \( V \) inside a conducting sphere is constant and equal to the potential at the surface of the sphere.
- The formula for the electric potential \( V \) at the surface of a sphere with charge \( Q \) and radius \( R \) is given by:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]
- Since the potential is the same throughout the sphere, the potential at the center is also:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]

4. **Final Results**:
- Therefore, the electric potential at the center of the sphere is:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]
- And the electric field at the center of the sphere is:
\[ E = 0 \, \text{N/C} \]

### Summary:
- Electric Potential at the center: \( \frac{Q}{4 \pi \epsilon_0 R} \)
- Electric Field at the center: \( 0 \, \text{N/C} \)