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

$\vec{p}_{x}=\int\limits_{0}^{l} \frac{\left(-\lambda_{0} x\right)}{l^{2}} x d x=-\lambda_{0} \frac{l}{3} \hat{i}$
$\vec{p}_{y}=\int\limits_{0}^{l} \frac{\left(+\lambda_{0} y\right)}{l^{2}} y d y=+\lambda_{0} \frac{l}{3} \hat{j}$
$\therefore \vec{p}=\frac{\lambda_{0} l}{3}(-\hat{i}+\hat{j}) ; \,\,\,\, |\vec{p}|=\frac{\sqrt{2}}{3} \lambda_{0} l$