Two charges of "5C" and "-2C" are situated at the points "(3a,0)" and "(-5a,0) respectively.The electric flux through a sphere of radius ' "4a'" having center at origin is:

Options

A. `(5)/(varepsilon_(0))`
B. `(7)/(varepsilon_(0))`
C. `(3)/(varepsilon_(0))`
D. `(-2)/(varepsilon_(0))`

Answer

To solve the problem, we need to find the electric flux through a sphere of radius \(4a\) centered at the origin, given two charges: \(q_1 = 5C\) located at \((3a, 0)\) and \(q_2 = -2C\) located at \((-5a, 0)\).

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Charge \(q_1 = 5C\) is located at \((3a, 0)\).
- Charge \(q_2 = -2C\) is located at \((-5a, 0)\).

2. **Determine the Sphere's Radius and Center**:
- The sphere has a radius of \(4a\) and is centered at the origin \((0, 0)\).

3. **Check Which Charges Are Inside the Sphere**:
- The distance of \(q_1\) from the origin is \(3a\). Since \(3a < 4a\), charge \(q_1\) is inside the sphere.
- The distance of \(q_2\) from the origin is \(5a\). Since \(5a > 4a\), charge \(q_2\) is outside the sphere.

4. **Apply Gauss's Law**:
- According to Gauss's law, the electric flux \(\Phi\) through a closed surface is given by:
\[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} \]
- Here, \(q_{\text{enclosed}}\) is the total charge inside the sphere. Since only \(q_1\) is inside:
\[ q_{\text{enclosed}} = q_1 = 5C \]

5. **Calculate the Electric Flux**:
- Substitute \(q_{\text{enclosed}}\) into Gauss's law:
\[ \Phi = \frac{5C}{\epsilon_0} \]

### Final Answer:
The electric flux through the sphere of radius \(4a\) is:
\[ \Phi = \frac{5}{\epsilon_0} \, \text{C/m}^2 \]