Question

An electric dipole is placed at distance x from an infinitely long rod of linear charge density $\lambda$. What is net force acting on dipole?

• A. $\frac{\lambda aq}{4\pi\epsilon_0 X^2}$
• B. $\frac{\lambda aq}{\pi\epsilon_0 x^2}$
• C. $\frac{\lambda aq}{2\pi\epsilon_0 x^2}$
• D. Zero

Answer 1

Answer: (B)

$\frac{\lambda aq}{\pi\epsilon_0 x^2}$

Answer 2

The electric field at a distance $r$ from an infinitely long rod with linear charge density $\lambda$ is $E(r) = \frac{\lambda}{2\pi\epsilon_0 r}$. For a dipole with charges $-q$ and $+q$ separated by $2a$, at distances $x-a$ and $x+a$ from the rod, the net force is: $F_{net} = (-q)E(x-a) + (+q)E(x+a)$ $F_{net} = -q \frac{\lambda}{2\pi\epsilon_0 (x-a)} + q \frac{\lambda}{2\pi\epsilon_0 (x+a)}$ $F_{net} = \frac{\lambda q}{2\pi\epsilon_0} \left( \frac{1}{x+a} - \frac{1}{x-a} \right)$ $F_{net} = \frac{\lambda q}{2\pi\epsilon_0} \left( \frac{(x-a) - (x+a)}{(x+a)(x-a)} \right) = \frac{\lambda q}{2\pi\epsilon_0} \left( \frac{-2a}{x^2 - a^2} \right)$ For $a \ll x$, $F_{net} \approx -\frac{2a\lambda q}{2\pi\epsilon_0 x^2} = -\frac{a\lambda q}{\pi\epsilon_0 x^2}$. The magnitude is $\frac{a\lambda q}{\pi\epsilon_0 x^2}$.