Electric Flux & Gauss's Law question from JEE Main 2023 (13 Apr, Shift 1)

Answer: 6

1. Between line and sheet, the line's field (away from line) and the sheet's field (away from sheet) are opposite: $E = \frac{\lambda}{2\pi\varepsilon_0 r} - \frac{\sigma}{2\varepsilon_0}$.
2. With $\lambda = 2\sigma$: $E = \frac{\sigma}{\pi\varepsilon_0 r} - \frac{\sigma}{2\varepsilon_0}$.
3. At P, $r = \frac{3}{\pi}$: $E_P = \left|\frac{\sigma}{3\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \frac{\sigma}{6\varepsilon_0}$.
4. At Q, $r = \frac{4}{\pi}$: $E_Q = \left|\frac{\sigma}{4\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \frac{\sigma}{4\varepsilon_0}$.
5. $\frac{E_P}{E_Q} = \frac{4}{6}$, so $a = 6$.