Question

A cube has sides of length $l=0.2\text{m}$. It is placed with one corner at the origin as shown in figure. The electric field is uniform and given by $\overrightarrow{E}=(2.5\hat{i}-4.2\hat{j})\text{N/C}$. Find the electric flux through the entire cube.

• A. $5{\text{N-m}}^2\text{/C}$
• B. $10{\text{N-m}}^2\text{/C}$
• C. $15{\text{N-m}}^2\text{/C}$
• D. Zero

Answer 1

The correct option is D Zero

Given:
$l=0.2\text{m};$
$\overrightarrow{E}=(2.5\hat{i}-4.2\hat{j})\text{N/C}$

We know that, electric flux through any surface is given by,
$\varphi =\overrightarrow{E}.\overrightarrow{A}$

For left surface (1) of cube,
$\overrightarrow{A_1}=(-0.04\hat{j}){\text{m}}^2[A=l^2={0.2}^2{\text{m}}^2=0.04{\text{m}}^2]$

So, ${\varphi }_1=\overrightarrow{E}.\overrightarrow{A_1}=(2.5\hat{i}-4.2\hat{j}).(-0.04\hat{j})=0.168{\text{N-m}}^2\text{/C}$

Similarly, for right surface (2),

${\varphi }_2=-0.168{\text{N-m}}^2\text{/C}$

for front surface (3),

${\varphi }_3=0.1{\text{N-m}}^2\text{/C}$

for back surface (4),

${\varphi }_4=-0.1{\text{N-m}}^2\text{/C}$

for top surface (5) of cube,
$\overrightarrow{A_5}=\left(0.04\hat{k}\right){\text{m}}^2$

So, ${\varphi }_5=\overrightarrow{E}.\overrightarrow{A_5}=(2.5\hat{i}-4.2\hat{j})\cdot \left(0.04\hat{k}\right)=0$

$\Rightarrow {\varphi }_5=0$

Similarly, for bottom surface (6),

${\varphi }_6=0$

Now, net electric flux through entire cube

${\varphi }_{\text{net}}={\varphi }_1+{\varphi }_2+{\varphi }_3+{\varphi }_4+{\varphi }_5+{\varphi }_6$

$\Rightarrow {\varphi }_{\text{net}}=0.168+(-0.168)+0.1+(-0.1)+0+0=0$

Hence, option $\left(d\right)$ is correct.