Question

An electric dipole is placed on x-axis in proximity to a line charge of linear charge density \(3.0 \times 10^{-6}\ \mathrm{C/m}\). Line charge is placed on z-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.

• A. \(0.485\ \mathrm{mC}\)
• B. \(815.1\ \mathrm{nC}\)
• C. \(8.8\ \mu\mathrm{C}\)
• D. \(4.44\ \mu\mathrm{C}\)

Hint:

The line charge field is \(E = \dfrac{\lambda}{2\pi\varepsilon_0 r}\); the two charges of the dipole lie on the same side of the origin and feel opposite-direction forces.

Answer 1

The Correct Option is D

Field of an infinite line charge: \(E(r) = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{2k\lambda}{r}\), with \(2k\lambda = 2\times 9\times10^{9}\times 3\times10^{-6} = 5.4\times10^{4}\).
Take both charges on the same side of the line charge (\(+q\) at 10 mm, \(-q\) at 12 mm), so the field has the same direction at both.
Force on \(+q\): \(qE_1 = q\times\dfrac{5.4\times10^4}{0.010} = 5.4\times10^{6}\,q\) (outward).
Force on \(-q\): \(qE_2 = q\times\dfrac{5.4\times10^4}{0.012} = 4.5\times10^{6}\,q\) (inward).
Net force \(= (5.4 - 4.5)\times10^{6}\,q = 0.9\times10^{6}\,q = 4\ \mathrm{N}\).
\(q = \dfrac{4}{0.9\times10^{6}} = 4.44\times10^{-6}\ \mathrm{C} = 4.44\ \mu\mathrm{C}\).
Answer: (4) \(4.44\ \mu\mathrm{C}\)