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A charge particle of mass m and charge q initally at rest is released in an electric field of magnitude E. Kinetic energy after time t will be -

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

\frac{Eqm}{2t}


Answers (1)

best_answer

Use newton's equation of motion -
\begin{array}{ll} v=u+a t & F=m a \\ v=0+a t & f=2 E \\ v=a t----(2) & m a=q E\\ \end{array}
                                            a=\frac{qE}{m}----\left ( 1 \right )
put the value of a in equation (1) we get -
v=\frac{q E t}{m}
kinetic energy k=\frac{1}{2} m \cdot u^2
\begin{aligned} & =\frac{1}{2} m \cdot\left(\frac{q E t}{m}\right)^2 \\ & =\frac{1}{2} m \cdot \frac{q^2 E^2 t^2}{m^2} \\ k \cdot E \cdot & =\frac{q^2 E^2 t^2}{2 m} \quad \text { } \end{aligned}
 

Posted by

Irshad Anwar

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