Physics - Elecrostatics Question with Solution | TestHub

Let's consider the following diagram:

The formula to calculate the electric field due to an infinitely long charged sheet is given by

$E_s=\frac{\sigma }{2{\epsilon }_0} ...\left(1\right)$

The formula to calculate the electric field due to an infinitely long line charge is given by

$E_l=\frac{\lambda }{2\pi {\epsilon }_{0}r} ...\left(2\right)$

Using equations (1) and (2), the electric field at $P$ can be calculated as follows:

$\begin{array}{rcl}{E}_P& =& \left|\frac{\sigma }{2{\epsilon }_0}-\frac{1}{4\pi {\epsilon }_0}\frac{2\lambda }{{\displaystyle \frac{3}{\pi }}}\right|\\ & =& \left|\frac{\sigma }{2{\epsilon }_0}-\frac{\lambda }{6{\epsilon }_0}\right|\\ & =& \frac{\sigma }{6{\epsilon }_0} ...\left(3\right)\end{array}$

Similarly, the electric field at $Q$ is given by

$\begin{array}{rcl}{E}_Q& =& \left|\frac{\sigma }{2{\epsilon }_0}-\frac{1}{4\pi {\epsilon }_0}\frac{2\lambda }{{\displaystyle \frac{4}{\pi }}}\right|\\ & =& \left|\frac{\sigma }{2{\epsilon }_0}-\frac{\lambda }{8{\epsilon }_0}\right|\\ & =& \frac{\sigma }{4{\epsilon }_0} ...\left(4\right)\end{array}$

Divide equation (3) by equation (4) to obtain the required ratio.

$\begin{array}{rcl}\frac{E_P}{E_Q}& =& \frac{{\displaystyle \frac{\sigma }{6{\epsilon }_0}}}{{\displaystyle \frac{\sigma }{4{\epsilon }_0}}}\\ & =& \frac{4}{6}\\ & & \end{array}$

Hence, $a=6$.