Question

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

• A. $60 \, ms^{-1}$
• B. $100 \, ms^{-1}$
• C. $90 \, ms^{-1}$
• D. $180 \, ms^{-1}$

Answer 1

Answer: (D)

$180 \, ms^{-1}$

Answer 2

Fixed charge $=2 \,mC =2 \times 10^{-3} C$ Mass of particle $=1 \,g =1 \times 10^{-3} kg$ Charge on particle $=1 \,\mu C =1 \times 10^{-6} C$ Distance between charges $=1 \,m$ Applying conservation of energy gives $U_{i}+K_{i} =U_{f}+K_{f}$ $\frac{K q_{1} \times q_{2}}{r_{1}}+\frac{1}{2} m v^{2}$ $=\frac{k q_{1} q_{2}}{r_{2}}+\frac{1}{2} m v^{2}$ $\frac{K q_{1} q_{2}}{r_{1}}=\frac{K q_{1} q_{2}}{r_{2}}+\frac{1}{2} m v^{2}$ $\therefore \frac{1}{2} m v^{2}=K q_{1} q_{2}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]$ $\Rightarrow K q_{1} q_{2} \left[\frac{r_{2}-r_{1}}{r_{1} r_{2}}\right]$ $\Rightarrow v^{2}=\frac{2 K q_{1} q_{2}\left(r_{2}-r_{1}\right)}{m r_{1} r_{2}}$ $\therefore v^{2}=\frac{2 \times 9 \times 10^{9} \times 10^{-6} \times 2 \times 10^{-3} \times(10-1)}{10^{-3} \times 1 \times 10}$ $v=180\, ms ^{-1}$