MHT CET · Physics · Electrostatics
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
- A \(\frac{\mathrm{q}}{\mathrm{dm} \mathrm{V}}\)
- B \(\frac{\mathrm{qm}}{\mathrm{Vd}}\)
- C \(\frac{\mathrm{qd}}{\mathrm{Vm}}\)
- D \(\frac{\mathrm{qV}}{\mathrm{dm}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{qV}}{\mathrm{dm}}\)
Step-by-step Solution
Detailed explanation
Electric field between the plates is \(\mathrm{E}=\left(\frac{\mathrm{V}}{\mathrm{d}}\right)\)
Force on the change can be written as
\(\begin{aligned}
& \mathrm{F}=\frac{\mathrm{qV}}{\mathrm{d}} \\
& \therefore \text { Acceleration }=\frac{\mathrm{F}}{\mathrm{m}}=\left(\frac{\mathrm{qV}}{\mathrm{md}}\right)
\end{aligned}\)
Force on the change can be written as
\(\begin{aligned}
& \mathrm{F}=\frac{\mathrm{qV}}{\mathrm{d}} \\
& \therefore \text { Acceleration }=\frac{\mathrm{F}}{\mathrm{m}}=\left(\frac{\mathrm{qV}}{\mathrm{md}}\right)
\end{aligned}\)
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