Question

A particle of mass $ 40 \,mg $ 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 $ 10\, cm $ from the fixed charge, it has a velocity of $ 50 \,cm/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
$\frac{1}{2}mv^{2}+\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Qq}{a} = \frac{1}{4\pi\varepsilon_{0}} \cdot\frac{Qq}{r}$
$ \frac{1}{2}\times 40 \times 10^{-6} \times \frac{1}{2}\times \frac{1}{2} = \frac{1}{4\pi\varepsilon_{0}}\cdot Qq\left[\frac{1}{r} - \frac{1}{a}\right] $
$\left(\because v = 50\, cms^{-1} = \frac{1}{2} \,ms^{-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}$
$ \times5 \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$