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

Answer

To solve the problem, we need to determine the electric potential and the 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 Charge Distribution**:
- When a conducting sphere is charged, the charge \( Q \) distributes itself uniformly on the outer surface of the sphere due to mutual repulsion between the charges.
- Inside the conducting material and at the center, there are no free charges.

2. **Electric Field Inside the Sphere**:
- According to Gauss's law, the electric field \( E \) inside a conductor in electrostatic equilibrium is zero.
- Since there are no charges inside the sphere, we can conclude that the electric field at the center of the sphere is:
\[ E = 0 \quad \text{(inside the conducting sphere)} \]

3. **Electric Potential Inside the Sphere**:
- The electric potential \( V \) inside a conductor is constant throughout its volume. This means that the potential at any point inside the conducting sphere, including the center, is the same as the potential at the surface of the sphere.
- The potential at the surface of a charged sphere is given by the formula:
\[ V = \frac{KQ}{R} \]
where \( K = \frac{1}{4\pi\epsilon_0} \).
- Therefore, the potential at the center of the sphere is:
\[ V = \frac{KQ}{R} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R} \]

4. **Final Results**:
- The electric field at the center of the conducting sphere is:
\[ E = 0 \]
- The electric potential at the center of the conducting sphere is:
\[ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R} \]

### Summary:
- Electric Field at the center: \( E = 0 \)
- Electric Potential at the center: \( V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R} \)