An electric dipole is placed on \({x}\)-axis in proximity to a line charge of linear charge density \(3.0 \times 10^{-6}\, {C} / {m}\). 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 \(4\, {N}\) is exerted on the dipole, find out the amount of positive or negative charge of the dipole.

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(D) \(4.44\, \mu {C}\)

Detailed explanation

\(r=10 \,{mm}, {x}=2\)

\(\left|\overrightarrow{{F}}_{{q}}\right|=\frac{2 {k} \lambda}{{r}} \cdot {q}\)

\(\left|\vec{F}_{-q}\right|=\frac{2 k \lambda}{r+x} \cdot q\)

\(\Rightarrow \quad\left|\vec{F}_{\text {net }}\right|=\frac{2 k \lambda q}{r}-\frac{2 k \lambda q}{r+x}\)

\(\left|\vec{F}_{n e t}\right|=\frac{2 k \lambda q \cdot x}{r(r+x)}\)

\(4=\frac{2 \times 9 \times 10^{9} \times 3 \times 10^{-6} \times q \times 2\, m m}{10 \,m m .12 \,m m}\)

\(\Rightarrow \quad q=4.44\, \mu C\)

Same subject

From JEE Mains