Q 12-01-155JEE MainJEE Main 2017 (9 Apr)Easy
Four closed surfaces and corresponding charge distributions are shown below. Let the respective electric fluxes through the surfaces be $\phi_1$, $\phi_2$, $\phi_3$ and $\phi_4$. Then:
Answer: (D) $\phi_1 = \phi_2 = \phi_3 = \phi_4$
By Gauss's law $\phi = q_{\text{enclosed}}/\varepsilon_0$; charges outside a surface do not contribute.
- $S_1$: $2q$
- $S_2$: $q + q + q - q = 2q$
- $S_3$: $q + q = 2q$
- $S_4$: $4q + 2q - 4q = 2q$ (the $3q$ is outside)
All four fluxes equal $2q/\varepsilon_0$.
Solution by Sreeraj P, M.Sc Physics