JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

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.

$(Use \frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N\cdotm^{2} / C^{2})$

Your answer

Solution

Linear charge density: $λ = \frac{Q}{L} = \frac{24 \times 1 0 ^{- 6}}{0.1} = 2.4 \times 1 0^{- 4} C/m$ .

Force on $q$ from an element $λ d x$ at distance $x$ : $d F = \frac{k q λ d x}{x ^{2}}$ . Integrate from $x = 0.02$ to $x = 0.12$ m:

$F = k q λ [- \frac{1}{x}]_{0.02}^{0.12} = k q λ (\frac{1}{0.02} - \frac{1}{0.12})$

$= (9 \times 1 0^{9}) (1 0^{- 6}) (2.4 \times 1 0^{- 4}) (50 - \frac{25}{3}) = 2.16 \times \frac{125}{3} = 90 N$ .

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