JEE Main 2023 — Physics Electromagnetism

Let's consider the following diagram:

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

$${E}_{s}=\frac{\sigma }{2{\epsilon }_{0}}...(1)$$Es​=2ϵ0​σ​...(1)

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

$${E}_{l}=\frac{\lambda }{2\pi {\epsilon }_{0}r}...(2)$$El​=2πϵ0​rλ​...(2)

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

$$\begin{matrix}{E}_{P} & = & |\frac{\sigma }{2{\epsilon }_{0}}-\frac{1}{4\pi {\epsilon }_{0}}\frac{2\lambda }{\frac{3}{\pi }}| \\ & = & |\frac{\sigma }{2{\epsilon }_{0}}-\frac{\lambda }{6{\epsilon }_{0}}| \\ & = & \frac{\sigma }{6{\epsilon }_{0}}...(3)\end{matrix}$$EP​​===​∣2ϵ0​σ​−4πϵ0​1​π3​2λ​∣∣2ϵ0​σ​−6ϵ0​λ​∣6ϵ0​σ​...(3)​

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

$$\begin{matrix}{E}_{Q} & = & |\frac{\sigma }{2{\epsilon }_{0}}-\frac{1}{4\pi {\epsilon }_{0}}\frac{2\lambda }{\frac{4}{\pi }}| \\ & = & |\frac{\sigma }{2{\epsilon }_{0}}-\frac{\lambda }{8{\epsilon }_{0}}| \\ & = & \frac{\sigma }{4{\epsilon }_{0}}...(4)\end{matrix}$$EQ​​===​∣2ϵ0​σ​−4πϵ0​1​π4​2λ​∣∣2ϵ0​σ​−8ϵ0​λ​∣4ϵ0​σ​...(4)​

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

$$\begin{matrix}\frac{{E}_{P}}{{E}_{Q}} & = & \frac{\frac{\sigma }{6{\epsilon }_{0}}}{\frac{\sigma }{4{\epsilon }_{0}}} \\ & = & \frac{4}{6} \\ & & \end{matrix}$$EQ​EP​​​==​4ϵ0​σ​6ϵ0​σ​​64​​

Hence, $a=6$a=6.