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 find 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 metallic spherical shell with a charge \( Q \) at its center. The inner radius of the shell is \( R_1 \) and the outer radius is \( R_2 \).
- The region we are interested in is the space between \( R_1 \) and \( R_2 \).

2. **Applying Gauss's Law**:
- According to Gauss's Law, the electric flux \( \Phi_E \) through a closed surface is proportional to the charge enclosed by that surface:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface.

3. **Choosing the Gaussian Surface**:
- We choose a spherical Gaussian surface of radius \( r \) such that \( R_1 < r < R_2 \).
- Since the charge \( Q \) is at the center, the entire charge \( Q \) is enclosed by this Gaussian surface.

4. **Calculating the Electric Field**:
- The electric field \( E \) at a distance \( r \) from the center is uniform over the Gaussian surface and points radially outward.
- The area \( A \) of the Gaussian surface is given by:
\[ A = 4\pi r^2 \]
- The electric flux through the Gaussian surface is:
\[ \Phi_E = E \cdot A = E \cdot 4\pi r^2 \]
- Setting the electric flux equal to the charge enclosed divided by \( \epsilon_0 \):
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
- Solving for \( E \):
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]

5. **Finding the Normal Component of the Electric Field**:
- The normal component of the electric field at any point on the Gaussian surface is the component of \( E \) that is perpendicular to the surface.
- Since the electric field \( E \) is directed radially outward and the normal to the surface is also radial, the angle between them is \( 0^\circ \).
- Therefore, the normal component of the electric field is:
\[ E_n = E \cdot \cos(0^\circ) = E \]
- However, we need to consider the electric field inside the conductor (metallic shell). Inside a conductor in electrostatic equilibrium, the electric field is zero.

6. **Conclusion**:
- Thus, the normal component of the electric field at any point on the Gaussian surface between \( R_1 \) and \( R_2 \) is:
\[ E_n = 0 \]

### Final Answer:
The normal component of the electric field at any point on the Gaussian surface with radius between \( R_1 \) and \( R_2 \) is **zero**.