A wire AB of length L has linear charge density `lambda = Kx`, where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for electric flux through the surface`.

Options

A. `(KL^2)/(2epsilon_0)`
B. `(KL)/(2epsilon_0)`
C. `(KL^2)/(epsilon_0)`
D. `(KL)/(epsilon_0)`

Explanation

To solve the problem of finding the electric flux through a Gaussian surface enclosing a wire with a linear charge density given by \(\lambda = Kx\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Linear Charge Density**:
The linear charge density is given as \(\lambda = Kx\), where \(x\) is the distance from point A along the wire. The wire extends from point A to point B, with a total length \(L\).

2. **Finding the Charge Element**:
Consider a small element of the wire of length \(dx\) at a distance \(x\) from point A. The charge \(dq\) on this small element can be expressed as:
\[ dq = \lambda \cdot dx = Kx \cdot dx \]

3. **Integrating to Find Total Charge**:
To find the total charge \(Q\) on the wire, we need to integrate \(dq\) from \(x = 0\) to \(x = L\):
\[ Q = \int_0^L dq = \int_0^L Kx \, dx \]
Performing the integration:
\[ Q = K \int_0^L x \, dx = K \left[ \frac{x^2}{2} \right]_0^L = K \cdot \frac{L^2}{2} = \frac{KL^2}{2} \]

4. **Applying Gauss's Law**:
According to Gauss's law, the electric flux \(\Phi\) through a closed surface is given by:
\[ \Phi = \frac{Q}{\epsilon_0} \]
Substituting the expression for \(Q\) we found:
\[ \Phi = \frac{\frac{KL^2}{2}}{\epsilon_0} = \frac{KL^2}{2\epsilon_0} \]

### Final Expression:
Thus, the electric flux through the Gaussian surface enclosing the wire is:
\[ \Phi = \frac{KL^2}{2\epsilon_0} \]