Question

The cube shown in the figure has sides of length $L=10.0\text{cm}$. The electric field is uniform, has a magnitude $E=4.00\times {10}^{-3}{\text{NC}}^{-1}$ and is parallel to the $xy$ plane at an angle of ${37}^{\circ }$ measured from the $+x$-axis toward the $+y$-axis.
Given $\sin {37}^{\circ }=0.6$, $\cos {37}^{\circ }=0.8$

$\left(ii\right)$ Electric flux passing through surface $S_1$ is

​​​​​​​$[$1 Mark$]$

• A. $-24\text{N}{\text{m}}^2{\text{C}}^{-1}$
• B. $42\text{N}{\text{m}}^2{\text{C}}^{-1}$
• C. $32\text{N}{\text{m}}^2{\text{C}}^{-1}$
• D. $-32\text{N}{\text{m}}^2{\text{C}}^{-1}$

Answer 1

The correct option is A $-24\text{N}{\text{m}}^2{\text{C}}^{-1}$

Given that
$E=4\times {10}^3\frac{N}{C}$
$A=100{\text{cm}}^2=100\times {10}^{-4}{\text{m}}^2={10}^{-2}{\text{m}}^2$
${\hat{n}}_{S_1}=-\hat{j}$ (Direction of area is outward normal)

We know
$\varphi =\overrightarrow{E}.\overrightarrow{A}=EA\cos \theta$ where $\overrightarrow{A}=A\hat{n}$

Angle between $\overrightarrow{A}$ and $\overrightarrow{E}$ is



$\therefore \theta ={90}^{\circ }+{37}^{\circ }$
$\varphi =EA\cos \theta$
$=4\times {10}^3\times {10}^{-2}\times \cos ({90}^{\circ }+{37}^{\circ })$
$=-40\sin {37}^{\circ }$
$=-40\times 0.6$
$=-24\text{N}{\text{m}}^2{\text{C}}^{-1}$
$\therefore$ option $\left(a\right)$ is correct.