Question

In free space, a particle A of charge $1\mu C$ is held fixed at a point P. Another particle B of the same charge and mass $4\mu g$ is kept at a distance of $1mm$ from P. if B is released, then its velocity at a distance of $9mm$ from P is :
[Take $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$]

• A. $2.0\times {10}^{3}m/s$
• B. $3.0\times {10}^{4}m/s$
• C. $1.5\times {10}^{2}m/s$
• D. $1.0m/s$

Answer 1

Answer: (A)

$2.0\times {10}^{3}m/s$

$W_E=-\left[\Delta U\right]=U_i-U_F=\frac{1}{2}m{v}^2$
$U=\frac{k{q}_1{q}_2}{r}$
$\frac{(9\times {10}^9)\times {10}^{-12}}{{10}^{-3}}-\frac{(9\times {10}^9)\times {10}^{-12}}{9\times {10}^{-3}}=\frac{1}{2}\times (4\times {10}^{-6})v^2$
$v^2=4\times {10}^6$
$v=2\times {10}^{3}m/s$