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

Options

A. 20 Vm
B. 10 Vm
C. 40 Vm
D. 5 Vm

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 by that surface. The formula for Gauss's Law is given by:

\[ \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]

where:
- \(\Phi\) is the electric flux through the surface,
- \(Q_{\text{enclosed}}\) is the total charge enclosed within the surface,
- \(\varepsilon_0\) is the permittivity of free space.

### Step-by-step Solution:

1. **Identify the Given Information:**
- The electric flux through a spherical surface of radius 10 cm (Φ₁) is given as \(20 \, \text{V} \cdot \text{m}\).
- We need to find the electric flux through a concentric spherical surface of radius 20 cm (Φ₂).

2. **Apply Gauss's Law to the Inner Sphere:**
- According to Gauss's Law, for the inner sphere (radius = 10 cm):
\[ \Phi_1 = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]
- Given that \(\Phi_1 = 20 \, \text{V} \cdot \text{m}\), we can express the enclosed charge \(Q_{\text{enclosed}}\) as:
\[ Q_{\text{enclosed}} = \Phi_1 \cdot \varepsilon_0 \]

3. **Apply Gauss's Law to the Outer Sphere:**
- For the outer sphere (radius = 20 cm), the charge enclosed remains the same since it is concentric and encloses the same volume:
\[ \Phi_2 = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]
- Since \(Q_{\text{enclosed}}\) is the same for both spheres, we can substitute:
\[ \Phi_2 = \Phi_1 \]

4. **Conclusion:**
- Therefore, the flux through the outer sphere (Φ₂) is also:
\[ \Phi_2 = 20 \, \text{V} \cdot \text{m} \]

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

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