A spherical conducting shell of radius `r_(0)` carry a charge `q_(0)`. Then value of electric field inside it is :-

Options

A. `kq_(0)//r^(2)`
B. `kq_(0)//r^(3)`
C. zero
D. uniform but non zero

Answer

To solve the problem of finding the electric field inside a spherical conducting shell of radius \( r_0 \) that carries a charge \( q_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Properties of Conductors**:
- A conducting shell allows charges to move freely. When a charge is placed on a conducting shell, it redistributes itself uniformly on the outer surface of the shell.

2. **Electric Field Inside a Conductor**:
- Inside a conductor in electrostatic equilibrium, the electric field is always zero. This is because any electric field would cause the free charges to move until they reach equilibrium.

3. **Applying Gauss's Law**:
- To mathematically confirm that the electric field inside the shell is zero, we can use Gauss's Law, which states:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface.

4. **Choosing a Gaussian Surface**:
- We choose a Gaussian surface that is a sphere of radius \( r \) where \( r < r_0 \) (inside the conducting shell).

5. **Calculating the Enclosed Charge**:
- Since all the charge \( q_0 \) resides on the surface of the conducting shell, the charge enclosed by our Gaussian surface (which is inside the shell) is zero:
\[ Q_{\text{enc}} = 0 \]

6. **Applying Gauss's Law**:
- Substituting \( Q_{\text{enc}} = 0 \) into Gauss's Law gives:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = 0 \]
- This implies that the electric field \( E \) inside the shell must also be zero, since the area \( dA \) is non-zero.

7. **Conclusion**:
- Therefore, the electric field inside the spherical conducting shell is:
\[ E = 0 \]

### Final Answer:
The value of the electric field inside the spherical conducting shell is \( E = 0 \).