Electric field due to each charge distribution

• For an infinite sheet with surface charge density $+\sigma$: $E_{\text{sheet}}=\frac{\sigma}{2\varepsilon_0}$ This is constant everywhere and directed away from the sheet.

• For an infinite line charge with linear charge density $+\lambda$: $E_{\text{line}}=\frac{\lambda}{2\pi\varepsilon_0 r}$ where $r$ is the perpendicular distance from the line.

Geometry of the arrangement

The line charge and sheet are parallel and separated by $5\,\text{m}$.

Points $P$ and $Q$ lie between the line and sheet, at distances from the line: $r_P=\frac{3}{\pi}\,\text{m}, \qquad r_Q=\frac{4}{\pi}\,\text{m}$ towards the sheet.

Since both charges are positive:

• Field due to the line at points between line and sheet is directed away from the line, i.e. toward the sheet.
• Field due to the sheet at points on the line side is directed away from the sheet, i.e. toward the line.

Hence, the two fields are opposite in direction.

Field at point $P$

Field due to line at $P$: $E_{L,P}=\frac{\lambda}{2\pi\varepsilon_0\left(\frac{3}{\pi}\right)}=\frac{\lambda}{6\varepsilon_0}$

Field due to sheet: $E_S=\frac{\sigma}{2\varepsilon_0}$

So resultant magnitude at $P$ is $E_P=\left|\frac{\lambda}{6\varepsilon_0}-\frac{\sigma}{2\varepsilon_0}\right|$

Field at point $Q$

Field due to line at $Q$: $E_{L,Q}=\frac{\lambda}{2\pi\varepsilon_0\left(\frac{4}{\pi}\right)}=\frac{\lambda}{8\varepsilon_0}$

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

Use the condition $2|\sigma|=|\lambda|$

Since both are given positive, we take $\lambda=2\sigma$

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

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

Find the ratio

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

Given $\frac{E_P}{E_Q}=\frac{4}{a}$

so, $\frac{4}{a}=\frac{2}{3}$ $a=6$

Final answer

$\boxed{6}$