A thin infinite sheet charge and an infinite line charge of respective charge densities $$+\sigma$$ and $$+\lambda$$ are placed parallel at $$5$$ 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. $$E_P$$ and $$E_Q$$ are the magnitudes of resultant electric field intensities at point P and Q respectively. If $$\frac{E_P}{E_Q} = \frac{4}{a}$$ for $$2|\sigma| = |\lambda|$$, then the value of $$a$$ is _____.

Given: An infinite sheet charge ($$+\sigma$$) and an infinite line charge ($$+\lambda$$) are parallel, separated by 5 m. Points P and Q are at distances $$\frac{3}{\pi}$$ m and $$\frac{4}{\pi}$$ m from the line charge, respectively, toward the sheet charge. Given $$2|\sigma| = |\lambda|$$.

Electric fields:

Sheet charge: $$E_{sheet} = \frac{\sigma}{2\varepsilon_0}$$ (constant, directed away from the sheet)

Line charge at distance $$r$$: $$E_{line} = \frac{\lambda}{2\pi\varepsilon_0 r}$$ (directed away from the line)

Both P and Q lie between the line and the sheet, so $$E_{line}$$ points toward the sheet and $$E_{sheet}$$ points toward the line (they oppose each other).

At point P (distance $$\frac{3}{\pi}$$ from line):

$$E_{line,P} = \frac{\lambda}{2\pi\varepsilon_0 \cdot \frac{3}{\pi}} = \frac{\lambda}{6\varepsilon_0}$$

$$E_P = \left|\frac{\lambda}{6\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right|$$

At point Q (distance $$\frac{4}{\pi}$$ from line):

$$E_{line,Q} = \frac{\lambda}{2\pi\varepsilon_0 \cdot \frac{4}{\pi}} = \frac{\lambda}{8\varepsilon_0}$$

$$E_Q = \left|\frac{\lambda}{8\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right|$$

Substituting $$\lambda = 2\sigma$$:

$$E_P = \left|\frac{2\sigma}{6\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \left|\frac{\sigma}{3\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \frac{\sigma}{6\varepsilon_0}$$

$$E_Q = \left|\frac{2\sigma}{8\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \left|\frac{\sigma}{4\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \frac{\sigma}{4\varepsilon_0}$$

The ratio:

$$\frac{E_P}{E_Q} = \frac{\sigma/(6\varepsilon_0)}{\sigma/(4\varepsilon_0)} = \frac{4}{6} = \frac{2}{3} = \frac{4}{6}$$

Comparing with $$\frac{E_P}{E_Q} = \frac{4}{a}$$:

$$a = 6$$

Therefore, the answer is $$\mathbf{6}$$.