`20 mu C` charge is placed inside a closed surface then flux related to surface is `phi`. If `80 mu C` charge is added inside the surface then charge in flux is :-

Options

A. `4 phi`
B. `5 phi`
C. `phi`
D. `8 phi`

Answer

To solve the problem, we will use Gauss's Law, which states that the electric flux (Φ) through a closed surface is directly proportional to the charge (q) enclosed within that surface. The relationship is given by:

\[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} \]

where:
- \( \Phi \) is the electric flux,
- \( q_{\text{enclosed}} \) is the total charge enclosed within the surface,
- \( \epsilon_0 \) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the initial charge and its flux**:
- Initially, there is a charge of \( q_1 = 20 \, \mu C \) inside the closed surface.
- According to Gauss's Law, the initial flux \( \Phi \) is given by:
\[ \Phi = \frac{q_1}{\epsilon_0} = \frac{20 \, \mu C}{\epsilon_0} \]

2. **Add the new charge**:
- An additional charge of \( q_2 = 80 \, \mu C \) is added inside the same closed surface.
- The total charge now enclosed by the surface is:
\[ q_{\text{total}} = q_1 + q_2 = 20 \, \mu C + 80 \, \mu C = 100 \, \mu C \]

3. **Calculate the new flux**:
- The new flux \( \Phi' \) after adding the charge is:
\[ \Phi' = \frac{q_{\text{total}}}{\epsilon_0} = \frac{100 \, \mu C}{\epsilon_0} \]

4. **Relate the new flux to the initial flux**:
- We can express the new flux in terms of the initial flux:
\[ \Phi' = \frac{100 \, \mu C}{\epsilon_0} = 5 \cdot \frac{20 \, \mu C}{\epsilon_0} = 5 \Phi \]
- Thus, the new flux is \( 5 \Phi \).

5. **Determine the change in flux**:
- The change in flux \( \Delta \Phi \) is given by:
\[ \Delta \Phi = \Phi' - \Phi = 5 \Phi - \Phi = 4 \Phi \]

### Final Answer:
The change in flux when \( 80 \, \mu C \) charge is added inside the surface is \( 4 \Phi \).