Question

The electric flux from a cube of edge $l$ is $\varphi$. If edge of the cube is made $2l$ and charge enclosed is halved, then the flux value will be

• A. $4\varphi$
• B. $2\varphi$
• C. $\frac{\varphi }{2}$
• D. $\varphi$

Answer 1

The correct option is C $\frac{\varphi }{2}$

Total electric flux through a closed surface is
$\varphi =\frac{q_{\text{enc}}}{{\epsilon }_0}$

where $q_{enc}$ is amount of charge enclosed within the closed surface.

According to the problem, the charge enclosed $q_{\text{enc}}$ is halved, so the new flux${\varphi }^{{}^{\prime }}$ becomes
${\varphi }^{\prime }=\frac{q_{\text{enc}}}{2{\epsilon }_0}=\frac{\varphi }{2}$

Hence, option (c) is the correct answer.