Question

Linear charge densities of the two rods are given as ${\lambda }_1=\frac{-{\lambda }_{0}x}{l^2}$ and ${\lambda }_2=\frac{+{\lambda }_{0}y}{l^2}.$ If the dipole moment of the system of rods is $\frac{\sqrt{m}{\lambda }_{0}l}{n}$, then find $(m+n)$.

Answer 1

Answer: 5

${\vec{p}}_x=\underset{0}{\overset{l}{\int }}\frac{\left(-{\lambda }_{0}x\right)}{l^2}xdx=-{\lambda }_0\frac{l}{3}\hat{i}$
${\vec{p}}_y=\underset{0}{\overset{l}{\int }}\frac{\left(+{\lambda }_{0}y\right)}{l^2}ydy=+{\lambda }_0\frac{l}{3}\hat{j}$
$\therefore \vec{p}=\frac{{\lambda }_{0}l}{3}(-\hat{i}+\hat{j});|\vec{p}|=\frac{\sqrt{2}}{3}{\lambda }_{0}l$