Electrostatics

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Question 7 of 311

A point charge q=1μCq=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=24μCQ=24 \mu \mathrm{C}, distributed uniformly along its length, as shown in figure. Force between qq and wire is ____\_\_\_\_ N. (Use : 14πϵo=9×109 N.m2/C2\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{~N}. \mathrm{m}^{2} / \mathrm{C}^{2})

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Answer: 90

Consider a small element of length dxdx on the wire at a distance xx from the point charge qq.
The linear charge density of the wire is λ=QL=24×10−60.1=240×10−6\lambda = \frac{Q}{L} = \frac{24 \times 10^{-6}}{0.1} = 240 \times 10^{-6} C/m.
The charge on the small element is dQ=λdxdQ = \lambda dx.
The electrostatic force between the point charge qq and the element dQdQ is dF=kqdQx2=kqλdxx2dF = \frac{k q dQ}{x^2} = \frac{k q \lambda dx}{x^2}.
The total force FF is obtained by integrating dFdF from x=ax = a to x=a+Lx = a + L, where a=2a = 2 cm and L=10L = 10 cm.
F=∫0.020.12kqλx2dx=kqλ[−1x]0.020.12F = \int_{0.02}^{0.12} \frac{k q \lambda}{x^2} dx = k q \lambda \left[ -\frac{1}{x} \right]_{0.02}^{0.12}
F=kqλ(10.02−10.12)=kqQL(0.12−0.020.02×0.12)F = k q \lambda \left( \frac{1}{0.02} - \frac{1}{0.12} \right) = k q \frac{Q}{L} \left( \frac{0.12 - 0.02}{0.02 \times 0.12} \right)
F=kqQLLa(a+L)=kqQa(a+L)F = \frac{k q Q}{L} \frac{L}{a(a+L)} = \frac{k q Q}{a(a+L)}
Substituting the given values: k=9×109k = 9 \times 10^9 N.m2^2/C2^2, q=1×10−6q = 1 \times 10^{-6} C, Q=24×10−6Q = 24 \times 10^{-6} C, a=0.02a = 0.02 m, and a+L=0.12a+L = 0.12 m.
F=9×109×1×10−6×24×10−60.02×0.12F = \frac{9 \times 10^9 \times 1 \times 10^{-6} \times 24 \times 10^{-6}}{0.02 \times 0.12}
F=9×24×10−30.0024=216×10−32.4×10−3F = \frac{9 \times 24 \times 10^{-3}}{0.0024} = \frac{216 \times 10^{-3}}{2.4 \times 10^{-3}}
F=2162.4=90F = \frac{216}{2.4} = 90 N.

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