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A charged particle of mass m and charge q is released from rest in an electric field of uniform strength E. The kinetic energy of the particle after a time t will be
A. \[\dfrac{{2{E^2}{t^2}}}{{mq}}\]
B. \[\dfrac{{{E^2}m{q^2}}}{{2{E^2}}}\]
C. \[\dfrac{{{q^2}{E^2}{t^2}}}{{2m}}\]
D. \[\dfrac{{qEm}}{{2t}}\]

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Last updated date: 29th Mar 2024
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MVSAT 2024
Answer
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Hint: Students should remember the equation of electric field E and the equation of kinetic energy in the form of linear momentum. Along with these Newton’s all three laws of motion are very important for the questions in which objects falling from a particular height for a particular time.

Complete step-by-step answer:
We know the equation of force,
\[F = q \times E\] …..(1)
and kinetic energy,
\[KE = \dfrac{{{p^2}}}{{2m}}\] …..(2)
Now, using Newton’s law we can write,
\[F = \dfrac{{dp}}{{dt}}\] …..(3)
By equating equations (1) and (3), we get
\[\dfrac{{dp}}{{dt}} = q \times E\]
\[dp = q \times Edt\]
On integrating above equation,
\[\int {dp} = \int\limits_0^t q \times Edt\]
\[p = qEt\] …..(4)
Squaring and adding equation (4) in equation (2) we get,
\[KE = \dfrac{{{q^2}{E^2}{t^2}}}{{2m}}\]
is a required solution.
Hence, a charged particle of mass m and charge q is released from rest in an electric field of uniform strength E. The kinetic energy of the particle after a time t will be
\[KE = \dfrac{{{q^2}{E^2}{t^2}}}{{2m}}\]
The correct option is C.

Note: Students should remember the formulae of force and kinetic energy in different forms. The integration and derivatives of basic functions are necessary in physics to solve problems containing dynamic functions.
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