Question

A charge $Q$ is uniformly distributed over a rod of length $l$ held perpendicular to the face of a hypothetical cube of side length $l$ with the centre of the cube at one end of the rod as shown in the figure below. Find the flux of the electric field through the entire surface of the cube.

• A. $\frac{Q}{2{\in }_0}$
• B. $\frac{Q}{4{\in }_0}$
• C. $\frac{Q}{{\in }_0}$
• D. $\frac{Q}{6{\in }_0}$

Answer 1

The correct option is A $\frac{Q}{2{\in }_0}$


Let the linear charge density of this rod be $\lambda$.

$\therefore$ From the definition, $\lambda =\frac{Q}{l}$

Since, the centre of the cube lies at one end of the rod and the cube is of side $l$, thus the length of rod inisde the cube is $\frac{l}{2}$.

So, the charge inside the cube is,

$q=\lambda \times \frac{l}{2}=\frac{Q}{l}\times \frac{l}{2}$

$\Rightarrow q=\frac{Q}{2}$

Now, by using Gauss's law,

$\varphi =\frac{q_{\text{enclosed}}}{{\in }_0}=\frac{Q}{2{\in }_0}$

Hence, option (a) is the correct answer.