A point charge is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge , distributed uniformly along its length, as shown in figure. Force between and wire is _______ N. [2026]

Question

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.

$(Use: \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N·m}^{2} / C^{2})$ [2026]

Smart Achievers · Accepted Answer

(90)

$F = \int d F = \int_{2 cm}^{12 cm} \frac{k q λ d x}{x^{2}} = k q λ (\frac{1}{2 \times 10^{- 2}} - \frac{1}{12 \times 10^{- 2}})$

$F = (9 \times 10^{9}) (10^{- 6}) (\frac{24 \times 10^{- 6}}{10^{- 1}}) (\frac{5}{12}) \times 10^{2}$

$= 9 \times 24 \times \frac{5}{12} = 90 N$

Solution

Ans.
(90)

$F = \int d F = \int_{2 cm}^{12 cm} \frac{k q λ d x}{x^{2}} = k q λ (\frac{1}{2 \times 10^{- 2}} - \frac{1}{12 \times 10^{- 2}})$

$F = (9 \times 10^{9}) (10^{- 6}) (\frac{24 \times 10^{- 6}}{10^{- 1}}) (\frac{5}{12}) \times 10^{2}$

$= 9 \times 24 \times \frac{5}{12} = 90 N$

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Solution

Q. A point charge q=1 μC is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge Q=24 μC, distributed uniformly along its length, as shown in figure. Force between q and wire is _______ N. (Use: 14πε0=9×109 N·m2/C2) [2026]

Ans. (90) F=∫dF=∫2 cm12 cmkqλ dxx2=kqλ(12×10-2-112×10-2) F=(9×109)(10-6)(24×10-610-1)(512)×102 =9×24×512=90 N

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Ans.
(90)

$F = \int d F = \int_{2 cm}^{12 cm} \frac{k q λ d x}{x^{2}} = k q λ (\frac{1}{2 \times 10^{- 2}} - \frac{1}{12 \times 10^{- 2}})$

$F = (9 \times 10^{9}) (10^{- 6}) (\frac{24 \times 10^{- 6}}{10^{- 1}}) (\frac{5}{12}) \times 10^{2}$

$= 9 \times 24 \times \frac{5}{12} = 90 N$