Question 640: Electrostatics & Capacitance

Question

A cubical volume is bounded by the surfaces $$\mathrm{x}=0, x=\mathrm{a}, y=0, y=\mathrm{a}, \mathrm{z}=0, z=\mathrm{a}$$. The electric field in the region is given by $$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i}$$. Where $$\mathrm{E}_{0}=4 	imes 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1}$$. If $$\mathrm{a}=2 \mathrm{~cm}$$, the charge contained in the cubical volume is $$\mathrm{Q} 	imes 10^{-14} \mathrm{C}$$. The value of $$\mathrm{Q}$$ is ________________.
(Take $$\epsilon_{0}=9 	imes 10^{-12} ~\mathrm{C}^{2} / \mathrm{Nm}^{2}$$)

Accepted Answer

$\begin{aligned} & \overrightarrow E  = {E_0}x\widehat i \\ & \phi_{\mathrm{net}}=\phi_{\mathrm{ABCD}}=\mathrm{E}_0 \mathrm{a} \cdot \mathrm{a}^2 \\ & \frac{\mathrm{q}_{\mathrm{en}}}{\in_0}=\mathrm{E}_0 \mathrm{a}^3 \\ & \mathrm{q}_{\mathrm{en}}=\mathrm{E}_0 \in_0 \mathrm{a}^3 \\ & =4 	imes 10^4 	imes 9 	imes 10^{-12} 	imes 8 	imes 10^{-6} \\ & =288 	imes 10^{-14} \mathrm{C} \\ & 	herefore \mathrm{Q}=288\end{aligned}$

Electrostatics & Capacitance

Integer / Numeric Answer Type