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(B) The electric field exists only in the region between the cylinders $(\sqrt{2}R  2R$,the

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$\frac{ Q }{60 \epsilon_0}$

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(B) The electric field exists only in the region between the cylinders $(\sqrt{2}R  2R$,the field is zero because the outer cylinder is grounded and encloses the inner charge.Using Gauss's Law for a cylinder of radius $r$ $(\sqrt{2}R The flux through an area element $dS = \ell dy$ on the plane is $d\phi = \vec{E} \cdot d\vec{S} = E \cos 	heta \ell dy$.From the geometry,$r = R \sec 	heta$ and $y = R 	an 	heta$,so $dy = R \sec^2 	heta d	heta$. Also $\cos 	heta = R/r$.Substituting these,$d\phi = \frac{\lambda}{2 \pi \kappa \epsilon_0 r} \cdot \frac{R}{r} \cdot \ell \cdot R \sec^2 	heta d	heta = \frac{\lambda \ell}{2 \pi \kappa \epsilon_0} d	heta$.The field is non-zero only for the segments $AB$ and $CD$ where $\sqrt{2}R For $r = \sqrt{2}R$,$\cos 	heta = R/(\sqrt{2}R) = 1/\sqrt{2} \Rightarrow 	heta = 45^\circ = \pi/4$.For $r = 2R$,$\cos 	heta = R/(2R) = 1/2 \Rightarrow 	heta = 60^\circ = \pi/3$.The flux through one segment is $\phi_{AB} = \int_{\pi/4}^{\pi/3} \frac{\lambda \ell}{2 \pi \kappa \epsilon_0} d	heta = \frac{Q}{2 \pi \kappa \epsilon_0} (\pi/3 - \pi/4) = \frac{Q}{2 \pi \kappa \epsilon_0} (\pi/12) = \frac{Q}{24 \kappa \epsilon_0}$.Given $\kappa = 5$,$\phi_{AB} = \frac{Q}{24 	imes 5 \epsilon_0} = \frac{Q}{120 \epsilon_0}$.Total flux through the plane is $\phi_{total} = \phi_{AB} + \phi_{CD} = 2 	imes \frac{Q}{120 \epsilon_0} = \frac{Q}{60 \epsilon_0}$.

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