Answer to Question #128545 in Physics for MUSAB

Question

A particle of mass 100 gm and charge 2 microcoulomb is released from rest at a distance of 50 cm from a fixed charge of 5 microcoulomb. Find the speed of the particle when its distance from the fixed charge become 3m?

Answer 1

$W=q_0(\phi_1-\phi_2)=q_0(k\frac{q}{r_1}-k\frac{q}{r_2})=kqq_0(\frac{1}{r_1}-\frac{1}{r_2})$

$W=mv^2/2\to v=\sqrt{2W/m}$

$v=\sqrt{2kqq_0(\frac{1}{r_1}-\frac{1}{r_2})/m}=$

$=\sqrt{2\cdot 9\cdot 10^9\cdot 5\cdot10^{-6} \cdot 2\cdot10^{-6}(\frac{1}{0.5}-\frac{1}{3})/0.1}=1.73(m/s)$