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 analyze the situation using Gauss's law, which relates the electric flux through a closed surface to the charge enclosed by that surface.

### Step-by-Step Solution:

1. **Understanding Electric Flux**: Electric flux (Φ) is defined as the product of the electric field (E) and the area (A) through which it passes, and it is given by the formula:
\[ \Phi = \int \vec{E} \cdot d\vec{A} \]
According to Gauss's law, the electric flux through a closed surface is proportional to the charge (Q) enclosed within that surface:
\[ \Phi = \frac{Q}{\epsilon_0} \]
where \(\epsilon_0\) is the permittivity of free space.

2. **Initial Condition**: The problem states that there is a charged body with an electric flux Φ associated with it. This means that there is a certain amount of charge Q enclosed by a surface around this charged body.

3. **Placing the Charged Body Inside a Metallic Container**: When the charged body is placed inside a metallic container, the container itself will respond to the electric field created by the charged body. However, since the container is metallic, it will redistribute its charges in such a way that the electric field inside the conductor becomes zero.

4. **Analyzing the Flux Outside the Container**: The key point is that the charge enclosed by the outer surface of the metallic container does not change. The total charge Q remains the same, as the metallic container does not add or remove any charge; it only redistributes it.

5. **Applying Gauss's Law to the Container**: For any Gaussian surface outside the metallic container, the charge enclosed is still Q (the charge of the body inside). Therefore, the electric flux Φ outside the container can be calculated using Gauss's law:
\[ \Phi_{\text{outside}} = \frac{Q}{\epsilon_0} \]
Since the charge Q has not changed, the flux outside the container remains equal to the original flux Φ.

6. **Conclusion**: The electric flux outside the metallic container will be equal to the electric flux associated with the charged body before it was placed inside the container.

Thus, the answer is:
\[ \Phi_{\text{outside}} = \Phi \]

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

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