JEE Main 2026 — Electrostatics Question with Solution

According to Gauss's law, the electric flux through a closed surface is given by $\Phi =\frac{q_{\text{enclosed}}}{{\epsilon }_0}$.

For the first sphere of radius $r_1=3$ cm centered at the origin, only the charge at $x=2$ cm is enclosed.

$q_{\text{enclosed},1}=8\mu \text{C}$

${\Phi }_1=\frac{8\mu \text{C}}{{\epsilon }_0}$

For the second sphere of radius $r_2=5$ cm centered at the origin, both charges at $x=2$ cm and $x=4$ cm are enclosed.

$q_{\text{enclosed},2}=8\mu \text{C}-2\mu \text{C}=6\mu \text{C}$

${\Phi }_2=\frac{6\mu \text{C}}{{\epsilon }_0}$

The ratio of the electric flux is:

$\frac{{\Phi }_1}{{\Phi }_2}=\frac{8}{6}=\frac{4}{3}$

Answer: $4:3$