Question

A charge particle of mass m and charge q is released from rest in an electric of constant magnitude E. The kinetic energy of the particle after time t is
$\begin{align}
  & \text{a)}\dfrac{{{\text{E}}^{\text{2}}}{{\text{q}}^{\text{2}}}{{\text{t}}^{\text{2}}}}{\text{2m}} \\
 & \text{b)}\dfrac{{{\text{E}}^{\text{2}}}{{\text{t}}^{\text{2}}}}{\text{mq}} \\
 & \text{c)}\dfrac{\text{Eqm}}{\text{2t}} \\
 & \text{d)}\dfrac{\text{E}{{\text{q}}^{\text{2}}}\text{m}}{\text{2}{{\text{t}}^{\text{2}}}} \\
\end{align}$

Answer 1

Hint: The above charge is kept in a constant electric field. Hence it will experience a constant force. Hence we can obtain the acceleration of the charge in the electric field. Further we will use this acceleration in the first kinematic equation and determine its velocity at any instant of time t. Since kinetic energy depends on the mass of the body (does not change with time when charge is kept in the electric field)and velocity, the velocity which we obtained above can be used to determine the kinetic energy.

Complete step by step answer:
The force on the charge in the above constant electric field is given by $\overline{F}=\overline{E}q$ where E is the electric field in which the charge is kept and q is the charge initially at rest let us say at time t=0. Since the electric field is constant the force F will not change with time. Hence as soon as the charge is introduced in the electric field initially at rest will undergo a constant acceleration. Since force equals mass times acceleration the magnitude of acceleration i.e. $\overline{a}=\dfrac{\overline{F}}{m}$, F is the constant force experienced by the charge and m is the mass of the charge. Now let us substitute the force due to the electric field in the above equation. Hence acceleration of the charge is given by $\overline{a}=\dfrac{\overline{E}q}{m}$.
Now let us use the first kinematic equation to determine the velocity of the charge at any instant time t for anybody under acceleration.
$\text{V- U=}\overline{\text{a}}\text{t}$ where V is the velocity of any body at time t, U is the initial velocity of any body at time t=0, a is the acceleration of the body under a constant force.
Now let substitute acceleration of the charge in first kinematic equation,
$\text{V- U=}\dfrac{\overline{E}q}{m}\text{t}$, now since U=0 i.e. initially the charge is at rest, the adjacent equation becomes
$\text{V=}\dfrac{\overline{E}q}{m}\text{t}$ is the velocity of the charge at any time t. Now let us substitute this in expression of kinetic energy of an object in motion. Mathematically kinetic energy of any body is given by $\text{K}\text{.E}=\dfrac{1}{2}m{{v}^{2}}$ where m is the mass of the body and v is the velocity of the object having kinetic energy at that particular instant of time.
Now let us substitute the instantaneous velocity of the charge obtained in the above expression of kinetic energy.
$\text{K}\text{.E}=\dfrac{1}{2}m{{v}^{2}}$
$\text{K}\text{.E}=\dfrac{1}{2}m{{\left( \dfrac{\overline{E}q}{m}\text{t} \right)}^{2}}=\dfrac{{{E}^{2}}{{q}^{2}}{{t}^{2}}}{2m}\text{J}$

So, the correct answer is “Option A”.

Note: It is not given in the question about the nature of the charge. If the charge is positive then the charge will move towards the direction of the applied electric field. If the charge is negative then it will move in the direction opposite to the applied electric field.