JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Electrostatics

By Gauss's law the flux through a closed surface depends only on the charge it encloses, $\varphi =q_{enc}/{\epsilon }_0$. The sphere of radius $3$ cm centred at the origin encloses only the $8\mu C$ charge at $x=2$ cm, so ${\varphi }_1=\frac{8\mu C}{{\epsilon }_0}$. The sphere of radius $5$ cm encloses both the $8\mu C$ charge and the $-2\mu C$ charge at $x=4$ cm, so ${\varphi }_2=\frac{(8-2)\mu C}{{\epsilon }_0}=\frac{6\mu C}{{\epsilon }_0}$. Their ratio is $\frac{{\varphi }_1}{{\varphi }_2}=\frac{8}{6}=\frac{4}{3}$, that is $4:3$.