Question

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

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

Answer 1

The correct option is C

$\frac{\mathrm{\varphi }}{2}$

The explanation for correct answer:

Option (C) $\frac{\mathrm{\varphi }}{2}$

Step 1 Given data:

The electric flux from a cube of edge $\mathrm{l}$ is $\mathrm{\varphi }$.

Now the edge of the cube is made $2\mathrm{l}$ and the charge enclosed is halved.

Step 2 Formula used:

The electric flux is the total number of electric field lines passing a given area in a unit of time.

Gauss Law states that the total electric flux($\mathrm{\varphi }$) out of a closed surface is equal to the charge($\mathrm{q}$) enclosed divided by the permittivity(${\mathrm{\epsilon }}_0$),

$\mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$

Step 3 Electric flux:

Here, $\mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$

So, the total electric flux($\mathrm{\varphi }$) is proportional to the charge($\mathrm{q}$) enclosed.

That is, if the charge enclosed is halved then the total electric flux also gets halved.

Therefore,

$$\begin{gathered} \mathrm{\varphi }=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\mathrm{\epsilon }}_0} \\ \mathrm{\varphi }=\frac{{\displaystyle \mathrm{q}}}{2{\mathrm{\epsilon }}_0} \\ \mathrm{\varphi }=\frac{1}{2}\left(\frac{{\displaystyle \mathrm{q}}}{{\mathrm{\epsilon }}_0}\right) \end{gathered}$$

Hence, The electric flux from a cube of edge $\mathrm{l}$ is $\mathrm{\varphi }$. If an edge of the cube is made $2\mathrm{l}$ and the charge enclosed is halved, its value will be $\frac{\mathrm{\varphi }}{2}$.