Question

A cube of side $a$ is placed such that the nearest face which is parallel to the y-z plane, is at a distance $a$ from the origin. The electric field components are $E_x=\beta x^{1/2},E_y=E_z=0$.
The charge within the cube is

• A. $\sqrt{2}\beta {ϵ}_a{a}^{5/2}C$
• B. $-\beta {ϵ}_o{a}^{5/2}C$
• C. $(\sqrt{2}-1)\beta {ϵ}_o{a}^{5/2}C$
• D. zero

Answer 1

The correct option is C $(\sqrt{2}-1)\beta {ϵ}_o{a}^{5/2}C$

Electric flux $\varphi =\int \overrightarrow{E}.\overrightarrow{dA}$

Electric flux will only be through the two faces perpendicular to the x-axis. (because the field is parallel to the other faces of the cube, hence $\varphi =\int \overrightarrow{E}.\overrightarrow{dA}=0$)


For the first face, $x=a$
Electric field at all points on this surface $E=\beta a^{1/2}$
Therefore, flux through this face ${\varphi }_1=\overrightarrow{E}.\overrightarrow{A}$
$=\beta a^{1/2}\hat{i}.a^2\left(\widehat{-i}\right)=\beta a^{1/2}.a^2(-1)$

For the second face, $x=2a$
Electric field at all points on this surface $E=\beta (2a{)}^{1/2}$
Therefore, flux through this face ${\varphi }_2=\overrightarrow{E}.\overrightarrow{A}$
$=\beta (2a{)}^{1/2}\hat{i}.a^2(\hat{i})=\beta (2a{)}^{1/2}.a^2$

Net flux $={\varphi }_1+{\varphi }_2=\beta a^{1/2}.a^2(-1)+\beta (2a{)}^{1/2}.a^2$
$=\beta a^{5/2}(\sqrt{2}-1)$
$=\frac{q}{{ϵ}_0}$ (Gauss's theorem)

Therefore, charge enclosed $q=(\sqrt{2}-1)\beta {ϵ}_o{a}^{5/2}C$