Two parallel plates separated by ' $\mathrm{d}$ ' $\mathrm{mm}$ are kept at potential difference of ' $V$ '…
Two parallel plates separated by ' $\mathrm{d}$ ' $\mathrm{mm}$ are kept at potential difference of ' $V$ ' volt. A particle of mass ' $m$ ' and charge ' $q$ ' enters in it with some velocity. The acceleration of the particle will be
$\frac{\mathrm{q}}{\mathrm{dm} \mathrm{V}}$
$\frac{\mathrm{qm}}{\mathrm{Vd}}$
$\frac{\mathrm{qd}}{\mathrm{Vm}}$
$\frac{\mathrm{qV}}{\mathrm{dm}}$
Solution
Electric field between the plates is $\mathrm{E}=\left(\frac{\mathrm{V}}{\mathrm{d}}\right)$
Force on the change can be written as
$\begin{aligned}
& \mathrm{F}=\frac{\mathrm{qV}}{\mathrm{d}} \\
& \therefore \text { Acceleration }=\frac{\mathrm{F}}{\mathrm{m}}=\left(\frac{\mathrm{qV}}{\mathrm{md}}\right)
\end{aligned}$