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(C) The electric field due to an infinite line charge is $E_L = \frac{\lambda}{2 \pi \varepsilon_0 r}$ and due to an infinite sheet charge is $E_S = \

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The electric field due to an infinite line charge is $E_L = \frac{\lambda}{2 \pi \varepsilon_0 r}$ and due to an infinite sheet charge is $E_S = \frac{\sigma}{2 \varepsilon_0}$.Since the points $P$ and $Q$ are between the line and the sheet,the fields are in opposite directions.At point $P$ $(r_P = \frac{3}{\pi} \ m)$: $E_P = \frac{\lambda}{2 \pi \varepsilon_0 (3/\pi)} - \frac{\sigma}{2 \varepsilon_0} = \frac{1}{2 \varepsilon_0} (\frac{\lambda}{3} - \sigma)$.At point $Q$ $(r_Q = \frac{4}{\pi} \ m)$: $E_Q = \frac{\lambda}{2 \pi \varepsilon_0 (4/\pi)} - \frac{\sigma}{2 \varepsilon_0} = \frac{1}{2 \varepsilon_0} (\frac{\lambda}{4} - \sigma)$.Given $2|\sigma| = |\lambda|$,we take $\lambda = 2\sigma$.$E_P = \frac{1}{2 \varepsilon_0} (\frac{2\sigma}{3} - \sigma) = \frac{1}{2 \varepsilon_0} (-\frac{\sigma}{3})$. Magnitude $|E_P| = \frac{\sigma}{6 \varepsilon_0}$.$E_Q = \frac{1}{2 \varepsilon_0} (\frac{2\sigma}{4} - \sigma) = \frac{1}{2 \varepsilon_0} (-\frac{\sigma}{2})$. Magnitude $|E_Q| = \frac{\sigma}{4 \varepsilon_0}$.$\frac{E_P}{E_Q} = \frac{\sigma / 6 \varepsilon_0}{\sigma / 4 \varepsilon_0} = \frac{4}{6}$.Comparing with $\frac{4}{a}$,we get $a = 6$.

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