A thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5 \mathrm{~m}$ distance from each other. Points 'P' and 'Q' are at $\frac{3}{\pi}$ m and $\frac{4}{\pi}$ m perpendicular distances from line charge towards sheet charge, respectively. '$\mathrm{E}_{\mathrm{P}}$' and '$\mathrm{E}_{\mathrm{Q}}$' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If $\frac{E_{p}}{E_{0}}=\frac{4}{a}$ for $2|\sigma|=|\lambda|$, then the value of $a$ is ___________.

With the given equations: <br/><br/> $$E_P = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{3 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{6 \varepsilon_0}\right| = \frac{\sigma}{6 \varepsilon_0}$$ <br/><br/> $$E_Q = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{4 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{8 \varepsilon_0}\right| = \frac{\sigma}{4 \varepsilon_0}$$ <br/><br/> Now we can find the ratio of $E_P$ to $E_Q$: <br/><br/> $$\frac{E_P}{E_Q} = \frac{\frac{\sigma}{6 \varepsilon_0}}{\frac{\sigma}{4 \varepsilon_0}} = \frac{2}{3}$$ <br/><br/> As given in the question, $\frac{E_P}{E_0} = \frac{4}{a}$, and since $\frac{E_P}{E_Q} = \frac{2}{3}$, we can say $\frac{E_P}{E_Q} = \frac{E_P}{2E_Q} = \frac{4}{2 \times 3}$ = $\frac{4}{ 6}$ . <br/><br/> So, the value of $a$ is $\boxed{6}$.