An infinitely long thin wire, having a uniform charge density per unit length of , is passing through a spherical shell of radius 1 m, as shown in the figure. A charge is distributed uniformly over the spherical shell. If the configuration of the charges

Question

An infinitely long thin wire, having a uniform charge density per unit length of $5 nC/m$ , is passing through a spherical shell of radius 1 m, as shown in the figure. A $10 nC$ charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points $P$ and $R$ , in Volt, is ________.

[Given: In SI units $\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$ , $\ln 2 = 0.7$ . Ignore the area pierced by the wire.] [2024]

Smart Achievers · Accepted Answer

(171)

$Due to wire, d V = - E . d x$

$\int_{V_{P}}^{V_{R}} d V = - \int_{0.5}^{2} \frac{2 k λ}{x} d x$

$V_{R} - V_{P} = - 2 k λ \ln (\frac{2}{0.5})$

$= - 2 \times 9 \times 10^{9} \times 3 \times 10^{- 9} \times 2 \times 0.7 = - 126 V$

$Due to sphere,$

$V_{R} - V_{P} = \frac{k Q}{2} - \frac{k Q}{1} = - \frac{k Q}{2}$

$= \frac{- 9 \times 10^{9} \times 10 \times 10^{- 9}}{2} = - 45 V$

$V_{R} - V_{P} = - 126 - 45 = - 171 V$

$or, V_{P} - V_{R} = 171 V$

Solution

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Solution

Ans. (171) Due to wire, dV=-E.dx ∫VPVRdV=-∫0.522kλxdx VR-VP=-2kλln(20.5) =-2×9×109×3×10-9×2×0.7=-126 V Due to sphere, VR-VP=kQ2-kQ1=-kQ2 =-9×109×10×10-92=-45 V VR-VP=-126-45=-171 V or, VP-VR=171 V

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