Question

A particle of mass $40mg$ and carrying a charge $5\times 10^{-9}C$ is moving towards a fixed charge of magnitude $10^{-8}C$ . When it is at a distance of $10cm$ from the fixed charge, it has a velocity of $50cm/s$ . At what distance from the fixed charge will the particle come momentarily to rest?

• A. $1.3\times 10^{-3}m$
• B. $1.9\times 10^{-3}m$
• C. $3.9\times 10^{-2}m$
• D. $4.7\times 10^{-2}m$

Answer 1

Answer: (D)

If the particle comes to rest momentarily at a distance $r$ from the fixed charge, then from conservation of energy, we have
$\frac{1}{2}m{v}^2+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Qq}{a}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Qq}{r}$
$\frac{1}{2}\times 40\times 10^{-6}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{4\pi {\epsilon }_0}\cdot Qq\left[\frac{1}{r}-\frac{1}{a}\right]$
$\left(\because v=50cm{s}^{-1}=\frac{1}{2}m{s}^{-1}\right)$
$=9\times 10^9\times 10^{-6}\times 5\times 10^{-9}\left[\frac{1}{r}-\frac{1}{10/100}\right]$
$\frac{1}{2}\times 40\times 10^{-6}\times \frac{1}{2}\times \frac{1}{2}=9\times 10^9\times 10^{-6}$
$\times 5\times 10^{-9}\left[\frac{1}{r}-10\right]$
$\frac{1}{r}-10=\frac{100}{9}$
$\frac{1}{r}=\frac{100}{9}+10=\frac{190}{9}$
$r=4.7\times 10^{-2}m$