A chance Q placed at thee center of a metallic spherical shell with inner and outer radii `R _1 and R_2` respectively . The normal component of the electric field at any point on the Gaussian surface with radius between `R_1and R_2` will be

Options

A. Zero
B. `Q/(4piR_(1)^(2))`
C. `Q/(4piR_(2)^(2))`
D. `Q/(4pi(R_1-R_2)^(2))`

Answer

To solve the problem, we need to analyze the situation involving a charge \( Q \) placed at the center of a metallic spherical shell with inner radius \( R_1 \) and outer radius \( R_2 \). We want to determine the normal component of the electric field at any point on a Gaussian surface with a radius between \( R_1 \) and \( R_2 \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charge \( Q \) located at the center of a metallic spherical shell.
- The shell has an inner radius \( R_1 \) and an outer radius \( R_2 \).
- We need to consider a Gaussian surface that lies between \( R_1 \) and \( R_2 \).

**Hint**: Visualize the charge and the spherical shell to understand the electric field distribution.

2. **Properties of Conductors**:
- In electrostatics, the electric field inside a conductor in electrostatic equilibrium is zero.
- Therefore, any electric field lines from the charge \( Q \) will induce charges on the inner surface of the shell.

**Hint**: Recall that electric field inside a conductor is zero when in electrostatic equilibrium.

3. **Induced Charges**:
- The positive charge \( Q \) at the center will induce a negative charge of \(-Q\) on the inner surface of the shell (at radius \( R_1 \)).
- This will leave a positive charge of \( +Q \) on the outer surface of the shell (at radius \( R_2 \)).

**Hint**: Remember that the total charge on a conductor must remain constant and that induced charges respond to the electric field.

4. **Applying Gauss's Law**:
- For a Gaussian surface between \( R_1 \) and \( R_2 \), the enclosed charge is zero because the inner surface has \(-Q\) and the outer surface has \( +Q\) but is outside our Gaussian surface.
- According to Gauss's Law, the electric flux through the Gaussian surface is given by:
\[ \Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- Since \( Q_{\text{enclosed}} = 0 \), we have:
\[ \Phi_E = 0 \]

**Hint**: Use Gauss's Law to relate electric flux to enclosed charge.

5. **Conclusion**:
- Since the electric flux is zero, the electric field \( E \) at any point on the Gaussian surface must also be zero.
- Therefore, the normal component of the electric field at any point on the Gaussian surface with radius between \( R_1 \) and \( R_2 \) is:
\[ E = 0 \]

**Final Answer**: The normal component of the electric field is \( 0 \).