Question

The cube as shown in Fig. has sides of length $L=10.0cm$. The electric field is uniform, has a magnitude $E=4.00\times {10}^{3}N{C}^{-1}$, and is parallel to the $xy-plane$ at an angle of ${37}^{\circ }$ measured from the $+x-axis$ towards the $+y-axis$.
Electric flux passing through surface $S_6$ is

• A. $24N{m}^2{C}^{-1}$
• B. $-24N{m}^2{C}^{-1}$
• C. $32N{m}^2{C}^{-1}$
• D. $-32N{m}^2{C}^{-1}$

Answer 1

Answer: (D)

$-32N{m}^2{C}^{-1}$

Here, the electric field is given by $\overrightarrow{E}=E\cos 37\hat{i}+E\sin 37\hat{j}$

Electric flux through back surface $S_6$ is $\varphi =\overrightarrow{E}.\hat{n}{S}_6$

Here $\hat{n}=-\hat{i}$, normal outward to the back surface $S_1=L^2$

Thus, $\varphi =(E\cos 37\hat{i}+E\sin 37\hat{j}).(-L^2\hat{i})=-E{L}^2\cos 37=-(4\times {10}^3)(0.1{)}^2\cos 37=-31.94\sim -32N{m}^2{C}^{-1}$