JEE Advanced · Physics · 17. Electrostatics
An infinitely long thin wire, having a uniform charge density per unit length of \(5 \mathrm{nC} / \mathrm{m}\), is passing through a spherical shell of radius \(1 \mathrm{~m}\), as shown in the figure. A \(10 \mathrm{nC}\) charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points \(\mathrm{P}\) and \(\mathrm{R}\), in Volt, is _______ .
[Given: In SI units \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9, \ln 2=0.7\). Ignore the area pierced by the wire.]

- A 158
- B 167
- C 171
- D 150
Answer & Solution
Correct Answer
(C) 171
Step-by-step Solution
Detailed explanation

due to wire
\(\mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dx}} \)
\( \int_{\mathrm{v}_{\mathrm{P}}}^{\mathrm{v}_{\mathrm{R}}} \mathrm{dV}=-\int_{0.5}^2 \frac{2 \mathrm{k} \lambda}{\mathrm{x}} \mathrm{dx} \)
\( \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-2 \mathrm{k} \lambda \ln \frac{2}{0.5} \)
\( =-2 \times 9 \times 10^9 \times 3 \times 10^{-9} \times 2 \times 0.7\) \(=-126 \mathrm{~V}\)
due to sphere
\(\mathrm{v}_{\mathrm{R}}-\mathrm{v}_P=\frac{\mathrm{kQ}}{2}-\frac{\mathrm{kQ}}{1}=-\frac{\mathrm{kQ}}{2}=\) \(\frac{-9 \times 10^9 \times 10 \times 10^{-9}}{2} \)
\(=-45 \mathrm{~V} \)
\( \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-126-45=-171 \mathrm{~V} \)
\( \mathrm{v}_{\mathrm{P}}-\mathrm{v}_{\mathrm{R}}=171 \mathrm{~V}\)
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