Electric charge are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25 V m. The flux over a concentric sphere of radius 20 cm will be

Options

A. 25 V m
B. 50 V m
C. 100 V m
D. 200 V m.

Answer

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

\[ \Phi = \frac{Q_{\text{in}}}{\epsilon_0} \]

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

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The flux through a spherical surface of radius 10 cm is given as \( \Phi_1 = 25 \, \text{V m} \).
- We need to find the flux through a concentric spherical surface of radius 20 cm, denoted as \( \Phi_2 \).

2. **Understand Gauss's Law**:
- According to Gauss's Law, the electric flux through a closed surface depends only on the charge enclosed within that surface, not on the size or shape of the surface.

3. **Apply Gauss's Law**:
- Since the charge is enclosed within both the 10 cm and 20 cm spherical surfaces, the total charge \( Q_{\text{in}} \) remains the same for both surfaces.
- Therefore, the flux through the first surface (10 cm) is equal to the flux through the second surface (20 cm):
\[ \Phi_2 = \Phi_1 \]

4. **Conclude the Calculation**:
- Since \( \Phi_1 = 25 \, \text{V m} \), we have:
\[ \Phi_2 = 25 \, \text{V m} \]

### Final Answer:
The flux over the concentric sphere of radius 20 cm will also be \( 25 \, \text{V m} \).