A charged body has an electric flux `phi` associated with it. The body is now placed inside a metallic container. The flux `phi`, outside the container will be

Options

A. Zero
B. Equal to `phi`
C. Greater than `phi`
D. Less than `phi`

Answer

To solve the problem, we need to understand the concept of electric flux and how it behaves when a charged body is placed inside a metallic container.

### Step-by-step Solution:

1. **Understanding Electric Flux**:
Electric flux (Φ) through a surface is defined as the total number of electric field lines passing through that surface. Mathematically, it is given by:
\[ \Phi = \frac{Q}{\varepsilon_0} \]
where \( Q \) is the charge enclosed and \( \varepsilon_0 \) is the permittivity of free space.

2. **Initial Condition**:
We have a charged body that has an electric flux \( \Phi \) associated with it. This means that if the charge \( Q \) is present, the flux through a closed surface surrounding the charge is:
\[ \Phi = \frac{Q}{\varepsilon_0} \]

3. **Placing the Charge Inside a Metallic Container**:
When the charged body is placed inside a metallic container, the metallic container will shield the electric field lines from escaping outside. However, the flux due to the charge inside the container does not change.

4. **Applying Gauss's Law**:
According to Gauss's Law, the electric flux through a closed surface is proportional to the charge enclosed within that surface. Since the charge \( Q \) is still inside the container, the total electric flux through the outer surface of the metallic container remains the same:
\[ \Phi_{\text{outside}} = \frac{Q}{\varepsilon_0} \]

5. **Conclusion**:
Therefore, the electric flux \( \Phi \) outside the metallic container will still be equal to the original flux associated with the charged body, which is:
\[ \Phi_{\text{outside}} = \Phi \]

### Final Answer:
The electric flux \( \Phi \) outside the container will be equal to \( \Phi \).

---