An infinitely long thin wire,having a uniform | vedclass

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(D) The total potential difference $V_P - V_R$ is the sum of the potential differences due to the wire and the spherical shell.$1$. Potential differen

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$171$

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(D) The total potential difference $V_P - V_R$ is the sum of the potential differences due to the wire and the spherical shell.$1$. Potential difference due to the infinitely long wire:The electric field at a distance $x$ from the wire is $E = \frac{2k\lambda}{x}$,where $k = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$ and $\lambda = 5 	imes 10^{-9} 	ext{ C/m}$.The potential difference $V_P - V_R = \int_{0.5}^{2} E \, dx = \int_{0.5}^{2} \frac{2k\lambda}{x} \, dx = 2k\lambda \ln\left(\frac{2}{0.5}ight) = 2k\lambda \ln(4) = 4k\lambda \ln(2)$.Substituting the values: $V_P - V_R = 4 	imes (9 	imes 10^9) 	imes (5 	imes 10^{-9}) 	imes 0.7 = 180 	imes 0.7 = 126 	ext{ V}$.$2$. Potential difference due to the spherical shell:Point $P$ is at distance $0.5 	ext{ m}$ (inside the shell of radius $R_s = 1 	ext{ m}$),and point $R$ is at distance $2 	ext{ m}$ (outside the shell).The potential inside a shell is constant: $V_P = \frac{kQ}{R_s} = \frac{(9 	imes 10^9) 	imes (10 	imes 10^{-9})}{1} = 90 	ext{ V}$.The potential outside a shell at distance $r$ is $V_R = \frac{kQ}{r} = \frac{(9 	imes 10^9) 	imes (10 	imes 10^{-9})}{2} = 45 	ext{ V}$.Thus,$V_P - V_R = 90 - 45 = 45 	ext{ V}$.Total potential difference: $V_P - V_R = 126 + 45 = 171 	ext{ V}$.

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