A charged body has an electric flux F assoclated with it. Now if the body is placed inside a conducting shell tthen the electric flux outside the shell is

Options

A. Zero
B. Greater then F
C. Less then F
D. Equal to F

Answer

To solve the question regarding the electric flux outside a conducting shell when a charged body is placed inside it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
Electric flux (Φ) through a surface is defined by the equation:
\[ \Phi = \frac{Q_{\text{in}}}{\varepsilon_0} \]
where \( Q_{\text{in}} \) is the charge enclosed by the surface and \( \varepsilon_0 \) is the permittivity of free space.

2. **Given Information**:
We are given that a charged body has an electric flux \( F \) associated with it. This implies:
\[ F = \frac{Q}{\varepsilon_0} \]
where \( Q \) is the charge of the body.

3. **Placing the Charged Body Inside a Conducting Shell**:
When the charged body is placed inside a conducting shell, the electric field inside the conductor must be zero. According to Gauss's law, if the electric field \( E \) is zero, then:
\[ \Phi = E \cdot A = 0 \]
This means that the net charge inside the conducting shell must also be zero.

4. **Induction of Charges**:
The charge \( Q \) on the charged body will induce a charge of \( -Q \) on the inner surface of the conducting shell. Consequently, a charge of \( +Q \) will appear on the outer surface of the conducting shell to maintain the neutrality of the shell.

5. **Calculating the Electric Flux Outside the Shell**:
To find the electric flux outside the conducting shell, we can consider a Gaussian surface that encloses the entire shell. The total charge enclosed by this Gaussian surface is \( +Q \) (the charge on the outer surface of the shell). Therefore, the electric flux through this Gaussian surface is given by:
\[ \Phi_{\text{outside}} = \frac{Q_{\text{outer}}}{\varepsilon_0} = \frac{Q}{\varepsilon_0} \]

6. **Relating Back to Given Flux**:
Since we established that \( F = \frac{Q}{\varepsilon_0} \), we can conclude that:
\[ \Phi_{\text{outside}} = F \]

### Final Answer:
The electric flux outside the conducting shell is \( F \).

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