Physics · Electrostatics

JEE Advanced 2024 — Paper 2 — Question 23

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 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 $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}, \ln 2=0.7$ . Ignore the area pierced by the wire.]

Question figure

Answer: 171

Numerical answer — enter this value.

Step-by-step solution

$\left(V_{P}-V_{R}\right)_{\text {line charge }}=2 k \lambda \ln \frac{r_{R}}{r_{P}}=126 \mathrm{~V}$

\begin{aligned} & \left(V_{P}-V_{R}\right)_{\text {sphere }}=k q\left(\frac{1}{1}-\frac{1}{r_{R}}\right)=\frac{k q}{2}=45 \mathrm{~V} \\& V_{P}-V_{R}=126+45=171 \mathrm{~V} . \end{aligned}

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Physics
Chapter: Electrostatics
Topic: Problems based on Application of Gauss's Law

A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle theta at P , as shown in the figure.…