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Physics · Electrostatics

JEE Main 2025 — 28 January, Morning Shift — Question 52

Three infinitely long wires with linear charge density λ are placed along the x -axis, y -axis and z axis respectively. Which of the following denotes an equipotential surface?

  1. Option A:

    xy+yz+zx=x y+y z+z x= constant

  2. Option B:

    (x+y)(y+z)(z+x)=(x+y)(y+z)(z+x)= constant

  3. Option C:

    (x2+y2)(y2+z2)(z2+x2)=\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)= constant

    Correct
  4. Option D:

    xyz=\mathrm{xyz}= constant

Answer: C

Step-by-step solution

v=−∫E→⋅dr→=∫2kλrdr=2kλln⁡r+c\quad \mathrm{v}=-\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=\int \frac{2 \mathrm{k} \lambda}{\mathrm{r}} \mathrm{dr}=2 \mathrm{k} \lambda \ln \mathrm{r}+\mathrm{c}

Net potential due to all wire

v=2kλln⁡x2+y2+2kλln⁡y2+z2+2kλln⁡z2+x2+c\mathrm{v}=2 \mathrm{k} \lambda \ln \sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}+2 \mathrm{k} \lambda \ln \sqrt{\mathrm{y}^{2}+\mathrm{z}^{2}}+2 \mathrm{k} \lambda \ln \sqrt{\mathrm{z}^{2}+\mathrm{x}^{2}}+\mathrm{c}

for v=cv=c

(x2+y2)(y2+z2)(z2+x2)=c\sqrt{\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)}=c

∴(x2+y2)(y2+z2)(z2+x2)=c\therefore\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left(\mathrm{y}^{2}+\mathrm{z}^{2}\right)\left(\mathrm{z}^{2}+\mathrm{x}^{2}\right)=\mathrm{c}

where c=\mathrm{c}= constant

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Potential Energy and Electric Potential

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