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

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Conducting Sphere**:
- A conducting sphere allows charges to move freely on its surface. When a charge \( Q \) is placed on the sphere, it distributes uniformly over the surface.

2. **Electric Field Inside the Conducting Sphere**:
- Inside a conducting sphere, the electric field \( E \) is zero. This is a fundamental property of conductors in electrostatic equilibrium. Therefore, at the center of the sphere, the electric field is:
\[ E_{\text{center}} = 0 \]

3. **Electric Potential Inside the Conducting Sphere**:
- The electric potential \( V \) inside a conductor is constant throughout its volume and equal to the potential at its surface. The potential at the surface of a conducting sphere with charge \( Q \) is given by the formula:
\[ V = \frac{KQ}{R} \]
where \( K = \frac{1}{4\pi \epsilon_0} \).

4. **Calculating the Electric Potential at the Center**:
- Since the electric potential is constant throughout the sphere, the potential at the center is the same as that at the surface:
\[ V_{\text{center}} = \frac{KQ}{R} \]

5. **Final Results**:
- Therefore, the electric potential at the center of the sphere is \( \frac{KQ}{R} \) and the electric field at the center of the sphere is \( 0 \).

### Summary of Results:
- Electric Potential at the center: \( V = \frac{KQ}{R} \)
- Electric Field at the center: \( E = 0 \)