Question

Electric field in the space is given as $\overrightarrow{E}=2x\hat{i}+y\hat{j}$, then charge enclosed inside cube of side length $2m$ as shown in figure will be

• A. $12{ϵ}_0$
• B. $24{ϵ}_0$
• C. $-6{ϵ}_0$
• D. $8{ϵ}_0$

Answer 1

The correct option is B $24{ϵ}_0$


Net flux associatated with cube is ${\varphi }_{net}={\varphi }_{ABCD}+{\varphi }_{EFGH}+$
${\varphi }_{ADEH}+{\varphi }_{BCFG}+{\varphi }_{ABGH}+{\varphi }_{CDEF}$
where, ${\varphi }_{BCFG}=E.A=2xA=2\left(2\right)A=4A$
${\varphi }_{EFGH}=E.A=yA=2A$
The flux through all other surfaces is zero.
${\varphi }_{net}=6A$
So, ${\varphi }_{net}=6A$ or ${\varphi }_{net}=\frac{q_{en}}{{ϵ}_0}$
$q_{en}=6A{ϵ}_0=24{ϵ}_0$
as, $A=2\times 2=4$