A point charge q =1 μ C is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge Q=24 μ C , distributed uniformly along its length, as shown in figure. Force between q and wire is \ \ \ \ N. (Use frac 1 4 π ε 0 =9 × 10 9 N . m 2 / C 2 )

Physics · Electrostatics

JEE Main 2026 — 24 January, Evening Shift — Question 45

A point charge $\mathrm{q}=1 \mu \mathrm{C}$ is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge $Q=24 \mu \mathrm{C}$ , distributed uniformly along its length, as shown in figure. Force between q and wire is $\_\_\_\_$ N. (Use $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} . \mathrm{m}^{2} / \mathrm{C}^{2}$ )

Answer: 90

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{F}=\int \mathrm{dF}=\int_{2 \mathrm{~cm}}^{12 \mathrm{~cm}} \frac{\mathrm{kq} \lambda \mathrm{dx}}{\mathrm{x}^{2}}=\mathrm{kq} \lambda\left(\frac{1}{2 \times 10^{-2}}-\frac{1}{12 \times 10^{-2}}\right)$ $\mathrm{F}=\left(9 \times 10^{9}\right)\left(10^{-6}\right)\left(\frac{24 \times 10^{-6}}{10^{-1}}\right)\left(\frac{5}{12}\right) \times 10^{2}$ $=9 \times 24 \times \frac{5}{12}=90 \mathrm{~N}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Field & motion of charge