Question

A line charge of length \( \frac{a}{2} \) is kept at the center of an edge BC of a cube ABCDEFGH having edge length \( a \). If the density of the line is \( \lambda C \) per unit length, then the total electric flux through all the faces of the cube will be : (Take \( \varepsilon_0 \) as the free space permittivity)

• A. \( \frac{\lambda a}{8 \varepsilon_0} \)
• B. \( \frac{\lambda a}{16 \varepsilon_0} \)
• C. \( \frac{\lambda a}{2 \varepsilon_0} \)
• D. \( \frac{\lambda a}{4 \varepsilon_0} \)

Hint:

When dealing with flux through symmetric shapes, use Gauss’s law to simplify the problem.

Answer 1

The Correct Option is A

The total charge enclosed within the cube is \( \frac{\lambda a}{4} \), as the charge is uniformly distributed along the edge BC. Using Gauss's law, the electric flux \( \Phi \) through the cube is: \[\phi = \frac{\lambda \frac{a}{2}}{4} = \frac{\lambda a}{8}\] \[ \Phi = \frac{q_{\text{in}}}{\varepsilon_0} = \frac{\lambda a}{8 \varepsilon_0} \] Thus, the total electric flux is \( \boxed{\frac{\lambda a}{8 \varepsilon_0}} \).