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

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C.
D.

Explanation

To solve the problem of how the electric field \( E \) varies with distance \( r \) from the center of a conducting shell carrying charge \(-Q\) with a point charge \( +Q \) placed at its center, we can follow these steps:

### Step 1: Understand the Setup
We have a conducting shell of radius \( R \) with a charge of \(-Q\) and a point charge \( +Q \) at the center. The electric field inside and outside the shell will be influenced by these charges.

### Step 2: Analyze the Electric Field Inside the Shell (0 < r < R)
- For a Gaussian surface inside the shell (where \( r < R \)), we consider the charge \( +Q \) at the center.
- According to Gauss's law, the electric flux through the Gaussian surface is given by:
\[ \Phi = \frac{Q_{\text{inside}}}{\epsilon_0} \]
where \( Q_{\text{inside}} = +Q \).

- The electric field \( E \) at a distance \( r \) from the center is uniform over the surface of the Gaussian sphere, so:
\[ \Phi = E \cdot 4\pi r^2 \]

- Setting the two expressions for flux equal gives:
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
Therefore, the electric field \( E \) at distance \( r \) is:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]
- This shows that the electric field inside the shell varies inversely with the square of the distance \( r \).

### Step 3: Analyze the Electric Field on the Surface of the Shell (r = R)
- At the surface of the shell (when \( r = R \)), the electric field is still given by the same formula:
\[ E = \frac{Q}{4\pi \epsilon_0 R^2} \]

### Step 4: Analyze the Electric Field Outside the Shell (r > R)
- For a Gaussian surface outside the shell (where \( r > R \)), we need to consider the total charge enclosed by the Gaussian surface.
- The total charge inside the Gaussian surface is:
\[ Q_{\text{total}} = +Q + (-Q) = 0 \]
- Therefore, according to Gauss's law:
\[ \Phi = \frac{Q_{\text{inside}}}{\epsilon_0} = 0 \]
- This implies that the electric field \( E \) outside the shell is:
\[ E = 0 \]

### Step 5: Summarize the Results
- For \( 0 < r < R \): \( E = \frac{Q}{4\pi \epsilon_0 r^2} \) (inversely proportional to \( r^2 \))
- For \( r = R \): \( E = \frac{Q}{4\pi \epsilon_0 R^2} \)
- For \( r > R \): \( E = 0 \)

### Final Answer
The electric field \( E \) varies with distance \( r \) as follows:
- It decreases with \( r^2 \) for \( 0 < r < R \),
- It is a constant value at \( r = R \),
- It becomes zero for \( r > R \).