Question 78 of 311

A thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5m$ distance from each other. Points $P$ and $Q$ are at $\dfrac{3}{\pi }m$ and $\dfrac{4}{\pi }m$ 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 $\dfrac{{E}_{P}}{{E}_{Q}}=\dfrac{4}{a}$ for $2|\sigma |=|\lambda |,$ then the value of $a$ is _____.

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Answer: 6

Let's consider the following diagram:
The formula to calculate the electric field due to an infinitely long charged sheet is given by
${E}_{s}=\dfrac{\sigma }{2{\epsilon }_{0}}...\left(1\right)$
The formula to calculate the electric field due to an infinitely long line charge is given by
${E}_{l}=\dfrac{\lambda }{2\pi {\epsilon }_{0}r}...\left(2\right)$
Using equations (1) and (2), the electric field at $P$ can be calculated as follows:
$\begin{matrix}{E}_{P} & = & \left|\dfrac{\sigma }{2{\epsilon }_{0}}-\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{2\lambda }{\dfrac{3}{\pi }}\right| \\ & = & \left|\dfrac{\sigma }{2{\epsilon }_{0}}-\dfrac{\lambda }{6{\epsilon }_{0}}\right| \\ & = & \dfrac{\sigma }{6{\epsilon }_{0}}...\left(3\right)\end{matrix}$
Similarly, the electric field at $Q$ is given by
$\begin{matrix}{E}_{Q} & = & \left|\dfrac{\sigma }{2{\epsilon }_{0}}-\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{2\lambda }{\dfrac{4}{\pi }}\right| \\ & = & \left|\dfrac{\sigma }{2{\epsilon }_{0}}-\dfrac{\lambda }{8{\epsilon }_{0}}\right| \\ & = & \dfrac{\sigma }{4{\epsilon }_{0}}...\left(4\right)\end{matrix}$
Divide equation (3) by equation (4) to obtain the required ratio.
$\begin{matrix}\dfrac{{E}_{P}}{{E}_{Q}} & = & \dfrac{\dfrac{\sigma }{6{\epsilon }_{0}}}{\dfrac{\sigma }{4{\epsilon }_{0}}} \\ & = & \dfrac{4}{6} \\ & & \end{matrix}$
Hence, $a=6$.