A point charge q = 1 is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge Q = 24, distributed uniformly along its length, as shown in figure. Force between q and wire is ______ N. (Use: (1)/(4πε₀) = 9 × 10⁹ N m²/C²) Numerical answer type. Work it out, then open the answer.

Answer: 90. Field on the axis of a rod at distance d from its near end: E = (kQ)/(d(d+L)), so F = (kqQ)/(d(d+L)). F = 9×10⁹ × 1×10⁻⁶× 24×10⁻⁶0.02×0.12 F = (0.216)/(0.0024) = 90 N So the force is 90 N.

A point charge q = 1 is located at a distance 2 cm from one end of a… | JEE Main 2026 (24 Jan, Shift 2)

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Electric Field & Field Lines

A point charge $q = 1\ \mu\text{C}$ is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge $Q = 24\ \mu\text{C}$ , distributed uniformly along its length, as shown in figure. Force between $q$ and wire is ______ N.

(Use: $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ N m}^2/\text{C}^2$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 90

Field on the axis of a rod at distance $d$ from its near end: $E = \dfrac{kQ}{d(d+L)}$ , so $F = \dfrac{kqQ}{d(d+L)}$ .
$F = \dfrac{9\times10^9 \times 1\times10^{-6}\times 24\times10^{-6}}{0.02\times0.12}$
$F = \dfrac{0.216}{0.0024} = 90\text{ N}$
So the force is 90 N.

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