Question

A small electric dipole having dipole moment $\overrightarrow{p}$ is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod placed along x axis, with one end coinciding with origin. If linear charge density of rod is $+\lambda$ and distance of dipole from rod is 'a', then calculate the electric force acting on dipole.

• A. $\frac{-P\lambda }{2\pi {ϵ}_0{a}^2}$
• B. $\frac{-3P\lambda }{4\pi {ϵ}_0{a}^2}$
• C. $\frac{-3P\lambda }{2\pi {ϵ}_0{a}^2}$
• D. $\frac{-P\lambda }{4\pi {ϵ}_0{a}^2}$

Answer 1

Answer: (D)

$\frac{-P\lambda }{4\pi {ϵ}_0{a}^2}$

Electric field at A due to dipole, $E_A=\frac{2KP}{r^3}$ where $K=\frac{1}{4\pi {ϵ}_o}$

$E_A=\frac{2KP}{(x+a{)}^3}$

Force at A due to dipole, $F_A=q{E}_A=(\lambda$ $dx)$ $\times \frac{2KP}{(x+a{)}^3}$

$F_A=\frac{2\lambda KP}{(x+a{)}^3}dx$

Total force exerted due to dipole on the rod, $F_D={\int }_0^{\infty }\frac{2\lambda KP}{(x+a{)}^3}dx$

$F_D=2\lambda KP\times \frac{1}{-2(x+a{)}^2}{|}_0^{\infty }$

$⟹F_A=\frac{\lambda KP}{a^2}$

Now force exerted by rod on dipole is equal and opposite to $F_D$ (Newton's law)

$\overrightarrow{F_R}=-F_D=\frac{-\lambda P}{4\pi {ϵ}_o{a}^2}$