Question

Electric potential in a region varies as $V=\left(1500x^{2}kVolt/m^2\right)$. Total electric charge lying inside the cube of side 10 cm shown in figure is $({ϵ}_0=8.9\times {10}^{-12}{C}^2/N{m}^2)$ :

• A. $-2.67\times {10}^{-8}C$
• B. $+2.67\times {10}^{-8}C$
• C. $-5.24\times {10}^{-8}C$
• D. $+5.24\times {10}^{-8}C$

Answer 1

Answer: (A)

$-2.67\times {10}^{-8}C$

Given, $V=1500x^2\frac{kvolt}{m^2}$

Electric field , $E=-\frac{\partial V}{\partial X}\hat{i}=3000x\frac{kvolt}{m}$

Flux,${\varphi }_E=\oint \overrightarrow{E}.\overrightarrow{ds}=-3000\times 0.1\times {0.1}^2\times {10}^{3}Vm$ (as $x=0.1m$ and $ds={0.1}^2$)

Using Gauss's law, $\Rightarrow q_{encl}={\varphi }_E.{ϵ}_0=-2.67\times {10}^{-8}C$ where ${ϵ}_0=8.9\times {10}^{-12}$