A conducting shell of radius R carries charge `-Q`. A point charge `+Q` is placed at the centre. The electric field E varies with distance `r` (from the centre of the shell) as

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the electric field \( E \) at different regions in relation to the conducting shell and the point charge. We will use Gauss's law to determine how the electric field varies with distance \( r \) from the center of the shell.

### Step-by-Step Solution:

1. **Identify the System**:
- We have a conducting shell of radius \( R \) with a charge of \( -Q \).
- A point charge \( +Q \) is placed at the center of the shell.

2. **Determine the Electric Field Inside the Shell**:
- For \( r < R \) (inside the conducting shell), we consider a Gaussian surface that is a sphere of radius \( r \) (where \( r < R \)).
- The only charge enclosed by this Gaussian surface is the point charge \( +Q \).
- According to Gauss's law:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \( Q_{\text{enc}} = +Q \), so:
\[ \Phi = \frac{Q}{\epsilon_0} \]
- The electric flux \( \Phi \) can also be expressed as:
\[ \Phi = E \cdot A = E \cdot 4\pi r^2 \]
- Setting these equal gives:
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
- Solving for \( E \):
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]
- Thus, the electric field inside the shell (for \( r < R \)) varies as:
\[ E \propto \frac{1}{r^2} \]

3. **Determine the Electric Field on the Surface of the Shell**:
- At \( r = R \), we can use the same formula:
\[ E = \frac{Q}{4\pi \epsilon_0 R^2} \]

4. **Determine the Electric Field Outside the Shell**:
- For \( r > R \), we consider a Gaussian surface outside the shell.
- The total charge enclosed by this Gaussian surface is \( +Q \) (from the point charge) and \( -Q \) (from the shell), resulting in a net charge of \( 0 \).
- Therefore, according to Gauss's law:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} = 0 \]
- This implies:
\[ E \cdot 4\pi r^2 = 0 \implies E = 0 \]
- Thus, for \( r > R \), the electric field is:
\[ E = 0 \]

5. **Summarize the Results**:
- For \( r < R \): \( E \propto \frac{1}{r^2} \)
- For \( r = R \): \( E = \frac{Q}{4\pi \epsilon_0 R^2} \)
- For \( r > R \): \( E = 0 \)

### Conclusion:
The electric field \( E \) varies with distance \( r \) as follows:
- It decreases with \( \frac{1}{r^2} \) for \( r < R \).
- It is constant at \( \frac{Q}{4\pi \epsilon_0 R^2} \) at \( r = R \).
- It becomes zero for \( r > R \).