Question

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

• A. $\sqrt{2}\alpha {\epsilon }_0{a}^{5/2}$
• B. $-\alpha {\epsilon }_0{a}^{5/2}$
• C. $(\sqrt{2}-1)\alpha {\epsilon }_0{a}^{5/2}$
• D. Zero

Answer 1

Answer: (C)

$(\sqrt{2}-1)\alpha {\epsilon }_0{a}^{5/2}$

${\varphi }_1=-E_1{a}^2=-\alpha a^{1/2}{a}^2=-\alpha a^{5/2}$
and ${\varphi }_2=E_2{a}^2=\alpha (2a{)}^{1/2}{a}^2=\sqrt{2}\alpha a^{5/2}$
${\varphi }_{net}=(\sqrt{2}-1)\alpha a^{5/2}$
Using the Gauss theorem, we get
$\varphi =\frac{q_{in}}{{ϵ}_0}$
$\Rightarrow q_{in}=\varphi {ϵ}_0=(\sqrt{2}-1){ϵ}_0\alpha a^{5/2}$.