Question

A particle of mass $1\text{kg}$ and charge $-2\mu \text{C}$(initially at a very large distance) is projected with initial velocity $200\text{m/s}$ towards a positive charged particle of charge $2\mu \text{C}$ which is fixed. The distance of $1\text{kg}$ particle from heavy particle when speed of $1\text{kg}$ particle is equal to twice of initial value is $\frac{x}{{10}^7}\text{m}$. Find the value of ‘$x$’.

Answer 1

Initial separation is large so, initial electrostatic potential energy is $0$
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m.(2u{)}^2-\left(\frac{1}{4\pi {ϵ}_0}\right)\frac{q^2}{r}$
$r=\frac{2q^2}{4\pi {ϵ}_0.3m{u}^2}=\frac{9\times {10}^9\times 2\times (2\times {10}^{-6}{)}^2}{3\times 1\times (200{)}^2}$
$r=\frac{6}{{10}^7}\text{m}$