Question

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

• A. $45{\text{ms}}^{-1}$
• B. $60{\text{ms}}^{-1}$
• C. $90{\text{ms}}^{-1}$
• D. $100{\text{ms}}^{-1}$

Answer 1

The correct option is C $90{\text{ms}}^{-1}$

The mass of the particle $m=2\text{gm}=2\times {10}^{-3}\text{kg}$

Charge of the particles $q_1=1\mu C$

The charge of the fixed particle , $q_2=1\text{mC}$

Initial distance $r_1=1\text{m}$

Final distance $r_2=10\text{m}$

Initial Total Energy $=$ Total Final Energy

$\therefore P.E_{initial}=P{E}_{Final}+K.E.$

$\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_1}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_2}+\frac{1}{2}m{v}^2$

$\frac{q_1{q}_2}{4\pi {\epsilon }_0}[\frac{1}{r_1}-\frac{1}{r_2}]=\frac{1}{2}m{v}^2$

Substituting the given data gives,

$9\times {10}^9\times {10}^{-6}\times {10}^{-3}[\frac{1}{1}-\frac{1}{10}]=\frac{1}{2}\times 2\times {10}^{-3}\times v^2$

$\frac{81\times {10}^3}{10}=v^2$

$\Rightarrow v=90{\text{ms}}^{-1}$