JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 26)
A thin infinite sheet charge and an infinite line charge of respective charge densities $$+\sigma$$ and $$+\lambda$$ are placed parallel at $$5 \mathrm{~m}$$ 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. '$$\mathrm{E}_{\mathrm{P}}$$' and '$$\mathrm{E}_{\mathrm{Q}}$$' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If $$\frac{E_{p}}{E_{0}}=\frac{4}{a}$$ for $$2|\sigma|=|\lambda|$$, then the value of $$a$$ is ___________.
Answer
6
Explanation
With the given equations:
$$E_P = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{3 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{6 \varepsilon_0}\right| = \frac{\sigma}{6 \varepsilon_0}$$
$$E_Q = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{4 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{8 \varepsilon_0}\right| = \frac{\sigma}{4 \varepsilon_0}$$
Now we can find the ratio of $$E_P$$ to $$E_Q$$:
$$\frac{E_P}{E_Q} = \frac{\frac{\sigma}{6 \varepsilon_0}}{\frac{\sigma}{4 \varepsilon_0}} = \frac{2}{3}$$
As given in the question, $$\frac{E_P}{E_0} = \frac{4}{a}$$, and since $$\frac{E_P}{E_Q} = \frac{2}{3}$$, we can say $$\frac{E_P}{E_Q} = \frac{E_P}{2E_Q} = \frac{4}{2 \times 3}$$ = $\frac{4}{ 6}$ .
So, the value of $$a$$ is $$\boxed{6}$$.
$$E_P = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{3 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{6 \varepsilon_0}\right| = \frac{\sigma}{6 \varepsilon_0}$$
$$E_Q = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{4 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{8 \varepsilon_0}\right| = \frac{\sigma}{4 \varepsilon_0}$$
Now we can find the ratio of $$E_P$$ to $$E_Q$$:
$$\frac{E_P}{E_Q} = \frac{\frac{\sigma}{6 \varepsilon_0}}{\frac{\sigma}{4 \varepsilon_0}} = \frac{2}{3}$$
As given in the question, $$\frac{E_P}{E_0} = \frac{4}{a}$$, and since $$\frac{E_P}{E_Q} = \frac{2}{3}$$, we can say $$\frac{E_P}{E_Q} = \frac{E_P}{2E_Q} = \frac{4}{2 \times 3}$$ = $\frac{4}{ 6}$ .
So, the value of $$a$$ is $$\boxed{6}$$.
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