Question

The electric field components in figure are $E_x=b\sqrt{x},E_y=E_z=0,$ Calculate the net flux through the cube.

• A. $(\sqrt{2}-1)b{a}^{\frac{5}{2}}$
• B. $b{a}^{\frac{5}{2}}$
• C. $\sqrt{b}{a}^{\frac{5}{2}}$
• D. $(\sqrt{5}-3)b{a}^{\frac{5}{2}}$

Answer 1

Answer: (A)

As $E_y=E_z=0$ so only faces along x-axis will contribute the flux through the cube.
Entering flux through the face at $x=a$ is ${\varphi }_{in}=E_x.a^2=b\sqrt{x}.a^2=b\sqrt{a}.a^2=b{a}^{5/2}$
Outgoing flux through the face at $x=2a$ is ${\varphi }_{out}=E_x.a^2=b\sqrt{2a}.a^2=b\sqrt{2a}.a^2=b{a}^{5/2}\sqrt{2}$
Thus , net flux through the cube is $\varphi ={\varphi }_{out}-{\varphi }_{in}=b{a}^{5/2}\sqrt{2}-b{a}^{5/2}=b{a}^{5/2}(\sqrt{2}-1)$