A particle of mass 2 g and charge $6 \mu \mathrm{C}$ is accelerated from rest through a potential difference…
A particle of mass 2 g and charge $6 \mu \mathrm{C}$ is accelerated from rest through a potential difference of 60 V . The speed acquired by the particle is
$0.6 \mathrm{~ms}^{-1}$
$1.2 \mathrm{~ms}^{-1}$
$1.8 \mathrm{~ms}^{-1}$
$0.3 \mathrm{~ms}^{-1}$
Solution
For the particle,
$\mathrm{m}=2 \mathrm{~g}, \mathrm{q}=6 \mu \mathrm{C}, \mathrm{~V}=60 \mathrm{~V}$
$\therefore$ The momentum of the particle is
$\begin{aligned}
& \mathrm{p}=\sqrt{2 \mathrm{mq}_{\mathrm{V}}} \Rightarrow \mathrm{mv}=\sqrt{2 \mathrm{mq}_{\mathrm{V}}} \\
& \therefore \text { Speed, } \mathrm{v}=\sqrt{\frac{2 \mathrm{q}_{\mathrm{V}}}{\mathrm{~m}}}=\sqrt{\frac{2 \times 6 \times 10^{-6} \times 60}{2 \times 10^{-3}}} \\
& =0.6 \mathrm{~ms}^{-1}
\end{aligned}$