Q 12-01-092JEE MainJEE Main 2023 (1 Feb, Shift 2)Medium
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 $\vec E=E_0x\hat i$, where $E_0=4\times10^4\ \text{N C}^{-1}\text{m}^{-1}$. If $a=2$ cm, the charge contained in the cubical volume is $Q\times10^{-14}$ C. The value of $Q$ is ______. (Take $\epsilon_0=9\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$)
Numerical value type. Enter your answer.
Answer: 288
Flux through $x=a$ is $E_0a\cdot a^2$; through $x=0$ it is zero.
$$q=\epsilon_0E_0a^3=9\times10^{-12}\times4\times10^4\times8\times10^{-6}=2.88\times10^{-12}\ \text{C}=288\times10^{-14}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics