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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2024 · Paper 2 · Physics · Electrostatics

An infinitely long thin wire, having a uniform charge density per...

An infinitely long thin wire, having a uniform charge density per unit length of 5nC/m5 \mathrm{nC} / \mathrm{m}, is passing through a spherical shell of radius 1 m1 \mathrm{~m}, as shown in the figure. A 10nC10 \mathrm{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\mathrm{P} and R\mathrm{R}, in Volt, is _______ . [Given: In SI units 14πϵ0=9×109,ln⁡2=0.7\frac{1}{4 \pi \epsilon_0}=9 \times 10^9, \ln 2=0.7. Ignore the area pierced by the wire.] An infinitely long thin wire, having a uniform charge density per unit length of 5 nC / m , is passing through a

Correct answer

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Step-by-step solution

An infinitely long thin wire, having a uniform charge density per unit length of 5 nC / m , is passing through a due to wire dV=−E→⋅dx→∫vPvRdV=−∫0.522kλxdxvR−vP=−2kλln⁡20.5=−2×9×109×3×10−9×2×0.7=−126 V\begin{aligned} & \mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dx}} \\ & \int_{\mathrm{v}_{\mathrm{P}}}^{\mathrm{v}_{\mathrm{R}}} \mathrm{dV}=-\int_{0.5}^2 \frac{2 \mathrm{k} \lambda}{\mathrm{x}} \mathrm{dx} \\ & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-2 \mathrm{k} \lambda \ln \frac{2}{0.5} \\ & =-2 \times 9 \times 10^9 \times 3 \times 10^{-9} \times 2 \times 0.7=-126 \mathrm{~V}\end{aligned} due to sphere vR−vP=kQ2−kQ1=−kQ2=−9×109×10×10−92=−45 VvR−vP=−126−45=−171 VvP−vR=171 V\begin{aligned} & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_P=\frac{\mathrm{kQ}}{2}-\frac{\mathrm{kQ}}{1}=-\frac{\mathrm{kQ}}{2}=\frac{-9 \times 10^9 \times 10 \times 10^{-9}}{2} \\ &=-45 \mathrm{~V} \\ & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-126-45=-171 \mathrm{~V} \\ & \mathrm{v}_{\mathrm{P}}-\mathrm{v}_{\mathrm{R}}=171 \mathrm{~V}\end{aligned}
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