Question

Charges are arranged to produce an electric field given by:
$E=(x,y,z)=cx\hat{i}+d\hat{j}$
where $c$ and $d$ are constants. How much charge is contained inside a cube with sides of length $a$ that has one corner at the origin and whose sides are parallel to the $x-,y-,$ and $z-$planes?

• A. $c{a}^2{ϵ}_0$
• B. $d{a}^3{ϵ}_0$
• C. $d{a}^2{ϵ}_0$
• D. $c{a}^3{ϵ}_0$

Answer 1

The correct option is D $c{a}^3{ϵ}_0$

$\overrightarrow{E}=cx\hat{i}+d\hat{j}$

$\varphi$ through all the sides.

${\varphi }_1$(ABCD) $=\overrightarrow{E}.d\overrightarrow{A}$

$=(cx\hat{i}+d\hat{j})\left(a^2\hat{j}\right)$

$=d{a}^2$

${\varphi }_2$ (BFGC) $=\overrightarrow{E}.d\overrightarrow{A}$

$=(cx\hat{i}+d\hat{j})\left(a^2\hat{j}\right)$

$=cx{a}^2{|}_{x=a}=c{a}^3$

${\varphi }_3$ (ADHE) $=\overrightarrow{E}.d\overrightarrow{A}$

$=(cx\hat{i}+d\hat{j})(a^2-\hat{i})$

$=-cx{a}^2{|}_{x=0}=0$

${\varphi }_4$ (EFGH) $\overrightarrow{E}.d\overrightarrow{A}=(cx\hat{i}+d\hat{j})(a^2-\hat{j})$

$=-d{a}^2$

${\varphi }_5$ (DCGH) $=(cx\hat{i}+d\hat{j})\left(a^2\hat{k}\right)=0$

${\varphi }_6$ (ABFE) $=(cx\hat{i}+d\hat{j})\left(a^2(-\hat{k})\right)=0$

Total flux through the cube $={\varphi }_1+{\varphi }_2+\cdot \cdot \cdot \cdot \cdot \cdot \cdot {\varphi }_6$

$=C{a}^3$

Now,

$\varphi =\frac{Q_{enc}}{{\epsilon }_0}$

$C{a}^3{\epsilon }_0=Q_{enc}$

Hence option D is correct