When a 10 `muC` charge is enclosed by a closed surface, the flux passing through the surface is `phi`. Now another 10 µC charge is placed inside the closed surface, then the flux passing through the surface is ______.

Options

A. `4phi`
B. `phi`
C. `2phi`
D. zero

Answer

To solve the problem step by step, 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:

\[ \Phi = \frac{Q}{\varepsilon_0} \]

where:
- Φ is the electric flux,
- Q is the total charge enclosed,
- ε₀ is the permittivity of free space (a constant).

### Step 1: Calculate the initial flux (Φ) with the first charge

When a 10 µC charge is enclosed by a closed surface, we can calculate the initial flux (Φ) using Gauss's law.

\[ \Phi = \frac{Q}{\varepsilon_0} = \frac{10 \times 10^{-6} \, \text{C}}{\varepsilon_0} \]

### Step 2: Add the second charge

Now, we introduce another charge of 10 µC inside the same closed surface. The total charge (Q_total) now becomes:

\[ Q_{\text{total}} = 10 \, \mu\text{C} + 10 \, \mu\text{C} = 20 \, \mu\text{C} \]

### Step 3: Calculate the new flux (Φ')

Using Gauss's law again, we can calculate the new flux (Φ') with the total charge:

\[ \Phi' = \frac{Q_{\text{total}}}{\varepsilon_0} = \frac{20 \times 10^{-6} \, \text{C}}{\varepsilon_0} \]

### Step 4: Relate the new flux to the initial flux

We can express the new flux in terms of the initial flux:

\[ \Phi' = 2 \cdot \Phi \]

This means that the new flux is twice the initial flux.

### Final Answer

Thus, the flux passing through the surface when both charges are present is:

\[ \Phi' = \frac{20 \, \mu\text{C}}{\varepsilon_0} \]