A line charge of length $\frac{\mathrm{a}}{2}$ is kept at the center of an edge BC of a cube ABCDEFGH having edge length 'a' as shown in the figure. If the density of line is λ C per unit length, then the total electric flux through all the faces of the cube will be  (Take, ${ϵ}_0$ as the free space permittivity)

We can solve this using Gauss's law, which states that the total electric flux $\Phi$ through a closed surface is proportional to the total charge enclosed within the surface. Mathematically:

$$\Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$

 Step 1: Determine the enclosed charge $Q_{\text{enclosed}}$

Assume three imaginary cubes (total four cubes) surrounding the edge BC. The given line charge has a linear charge density $\lambda$ (charge per unit length) and length $\frac{a}{2}$. Therefore, the total charge $Q_{\text{enclosed}}$ on the line is:

$$Q_{\text{enclosed}} = \lambda \cdot \frac{a}{2}$$

Step 2: Apply Gauss's law

The total electric flux through all the four cubes is:

$$\Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$

$$\Phi = \frac{\lambda \cdot \frac{a}{2}}{\epsilon_0}$$

 $$\Phi = \frac{\lambda a}{2 \epsilon_0}$$ 

Final Answer:

The total electric flux through one cube is:

${\mathrm{\Phi }}_{\mathrm{single} \mathrm{cube}}=\frac{\mathrm{\Phi }}{4}=\frac{\frac{\mathrm{\lambda a}}{2{\mathrm{ϵ}}_0}}{4}=\frac{\mathrm{\lambda a}}{8{\mathrm{ϵ}}_0}$