Question

A charge of mass $50\text{mg}$ and carrying $5\times {10}^{-9}\text{C}$ is approaching a fixed of ${10}^{-8}\text{C}$ with a velocity of $0.5{\text{ms}}^{-1}$. The distance of closest approach of the charge is

• A. $1.2\text{cm}$
• B. $4.2\text{cm}$
• C. $7.2\text{cm}$
• D. $8.6\text{cm}$

Answer 1

The correct option is C $7.2\text{cm}$

According to law of conservation of energy,

$\text{Loss of KE}=\text{Gain in PE}$

$\frac{1}{2}m{u}^2=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r}$

$\frac{1}{2}\times 50\times {10}^{-3}\times {10}^{-3}\times (0.5{)}^2=9\times {10}^9\times \frac{5\times {10}^{-9}\times {10}^{-8}}{r}$

$r=\frac{9\times 5}{25\times 25}=\frac{9}{125}\text{m}$

$=0.072\text{m}$

$r=7.2\text{cm}$