Question

A thin rod of length $2a$ is placed along $y$ -axis in the $x-y$ plane. The rod carries a charge density $\lambda$ as shown in figure. If point $P_1$ is located at $\left(0,2a\right)$ and $P_2$ at $\left(x,0\right),$ then find ${\left(\frac{x}{a}\right)}^2$ if potential at $P_1$ and $P_2$ are equal.

Answer 1

Answer: 3

$v_1=k{\int }_{-a}^a\frac{\lambda dy}{\left(2a-y\right)}=k\lambda ln3$
$v_2=k{\int }_{-a}^{+a}\frac{\lambda dy}{\sqrt{y^2+x^2}}$
$=2k\lambda ln\left(\frac{a+\sqrt{a^2+x^2}}{x}\right)$
$\because v_1=v_2$
$k\lambda ln3=2k\lambda ln\left(\frac{a+\sqrt{a^2+x^2}}{x}\right)$
$\sqrt{3}x=a+\sqrt{a^2+x^2}$
$\frac{x}{a}=\sqrt{3}=1.732$
$1.73$