Question

A particle of mass $1gm$ and charge $1\mu C$ is held at rest on a frictionless horizontal surface at distance $1m$ from the fixed charge $2mC$. If the particle is released, it will be repelled. The speed of the particle when it is at a distance of $10m$ from the fixed charge

• A. $60m{s}^{-1}$
• B. $100m{s}^{-1}$
• C. $90m{s}^{-1}$
• D. $180m{s}^{-1}$

Answer 1

Answer: (D)

Fixed charge $=2mC=2\times 10^{-3}C$
Mass of particle $=1g=1\times 10^{-3}kg$
Charge on particle $=1\mu C=1\times 10^{-6}C$
Distance between charges $=1m$
Applying conservation of energy gives
$U_i+K_i=U_f+K_f$
$\frac{K{q}_1\times q_2}{r_1}+\frac{1}{2}m{v}^2$
$=\frac{k{q}_1{q}_2}{r_2}+\frac{1}{2}m{v}^2$
$\frac{K{q}_1{q}_2}{r_1}=\frac{K{q}_1{q}_2}{r_2}+\frac{1}{2}m{v}^2$
$\therefore \frac{1}{2}m{v}^2=K{q}_1{q}_2\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$
$\Rightarrow K{q}_1{q}_2\left[\frac{r_2-r_1}{r_1{r}_2}\right]$
$\Rightarrow v^2=\frac{2K{q}_1{q}_2\left(r_2-r_1\right)}{m{r}_1{r}_2}$
$\therefore v^2=\frac{2\times 9\times 10^9\times 10^{-6}\times 2\times 10^{-3}\times (10-1)}{10^{-3}\times 1\times 10}$
$v=180m{s}^{-1}$