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QJEEM23Ph6a9b: JEE Main 2023 Physics - Class 12 JEEM23Ph6a9b Solution: A thin infinite sheet charge and an infinite line charge of…

A thin infinite sheet charge and an infinite line charge of respective charge densities +σ+\sigma and +λ+\lambda are placed parallel at 5 m5 \mathrm{~m} distance from each other. Points 'P' and 'Q' are at 3π\frac{3}{\pi} m and 4π\frac{4}{\pi} m perpendicular distances from line charge towards sheet charge, respectively. ' EP\mathrm{E}_{\mathrm{P}} ' and ' EQ\mathrm{E}_{\mathrm{Q}} ' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If EpE0=4a\frac{E_{p}}{E_{0}}=\frac{4}{a} for 2∣σ∣=∣λ∣2|\sigma|=|\lambda| , then the value of aa is ___________.

Step-by-Step Verified Solution

With the given equations:

EP=∣σ2ε0−14πε02λ3/π∣=∣σ2ε0−λ6ε0∣=σ6ε0E_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}

EQ=∣σ2ε0−14πε02λ4/π∣=∣σ2ε0−λ8ε0∣=σ4ε0E_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 EPE_P to EQE_Q :

EPEQ=σ6ε0σ4ε0=23\frac{E_P}{E_Q} = \frac{\frac{\sigma}{6 \varepsilon_0}}{\frac{\sigma}{4 \varepsilon_0}} = \frac{2}{3}

As given in the question, EPE0=4a\frac{E_P}{E_0} = \frac{4}{a} , and since EPEQ=23\frac{E_P}{E_Q} = \frac{2}{3} , we can say EPEQ=EP2EQ=42×3\frac{E_P}{E_Q} = \frac{E_P}{2E_Q} = \frac{4}{2 \times 3} = 46\frac{4}{ 6} .

So, the value of aa is 6\boxed{6} .

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