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  • Help me solve this A charged particle of mass mand charge qis released from rest in a uniform electric field E .Neglecting the effect of gravity, the kinetic energy of the charged particle after ‘t’ second is

 A charged particle of mass  m  and charge q  is released from rest in a uniform electric field  E . Neglecting the effect of gravity, the kinetic energy of the charged particle after ‘t’ second is

  • Option 1)

    \frac{Eq^2m}{2t^2}

  • Option 2)

    \frac{2E^2t^2}{mq}

  • Option 3)

    \frac{E^2q^2t^2}{2m}

  • Option 4)

    \frac{Eqm}{t}

 

Answers (1)

best_answer

As we have learned

when Charged Particle at rest in uniform field -

W=Q\cdot \Delta v

-When charge q is released in uniform electric field  then its acceleration a=\frac{qE}{m}  (is constant)

So its motion will be uniformly accelerated motion and its velocity after time t is given by

v=at=\frac{qE}{m}t \Rightarrow KE= 1/2mv^2= 1/2(\frac{qE}{m})t= \frac{q^2E^2t^2}{2m}

 

Option 1)

\frac{Eq^2m}{2t^2}

Option 2)

\frac{2E^2t^2}{mq}

Option 3)

\frac{E^2q^2t^2}{2m}

Option 4)

\frac{Eqm}{t}

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Plabita

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