Electrostatics Question 21
Question 21 - 01 February - Shift 2
A cubical volume is bounded by the surfaces $x=0, x=a, y=0, y=a, z=0, z=a$. The electric
field in the region is given by $\overrightarrow{{}E}=E_0 \times \hat{\dot{i}}$. Where $E_0=4 \times 10^{4} NC^{-1} m^{-1}$. If $a=2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is
Take $\in_0=9 \times 10^{-12} C^{2} / Nm^{2}$ )
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Answer: (288)
Solution:
Formula: Electric Flux
$ \phi _{net}=\phi _{ABCD}=E_0 a \cdot a^{2} $
$ \frac{q _{\text{en }}}{\epsilon_0}=E_0 a^{3} $
$q _{en}=E_0 \in_0 a^{3}$
$=4 \times 10^{4} \times 9 \times 10^{-12} \times 8 \times 10^{-6}$
$=288 \times 10^{-14} C$
$Q=288$
Ans. 288
