Question

An electron moves at a distance of 6cm when accelerated from rest by an electric field of strength$2\times {{10}^{4}}N{{C}^{-1}}$. Calculate the time of travel (mass of electron=$9.1\times {{10}^{-31}}Kg$).

Answer 1

Hint: The question is based on the motion of a uniform electric field. Under the influence of the electric field, electrons experience a force. Use the relation of force and electric field with charges. To solve further use a kinematic equation.

Complete step by step answer:
Electrons are under the influence of an electric field then there must be force which is exerting on that electron. Which you can understand by given formula,
$\begin{aligned} & \text{electric field=}\dfrac{\text{force}}{\text{charge}} \\\ & i.e. \\\ & E=\dfrac{F}{q} \\\ \end{aligned}$
$F=Eq$
Then according to Newton's second law of motion force is equal to product of mass and acceleration.
$F=ma$
Equating value of force (F)
We get,
$Eq=ma$
Where m= mass of electron
a = acceleration of electron
E = electric field
q= charge on electron
Therefore value of a is
$a=\dfrac{qE}{m}$
Now use kinematic equation,
$s=ut+\dfrac{a{{t}^{2}}}{2}$
At initial stage speed was zero so,
$u=0$
$s=\dfrac{a{{t}^{2}}}{2}$
But we want the value of time (t).
Rearrange the above equation, we get
$\begin{aligned} & t=\sqrt{\dfrac{2s}{a}} \\\ & t=\sqrt{\dfrac{2sm}{qE}} \\\ & t=\sqrt{\dfrac{2\times 6\times {{10}^{-2}}\times 9.1\times {{10}^{-31}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-4}}}} \\\ & t=\sqrt{\dfrac{34.12\times {{10}^{-10}}}{1}} \\\ & t=5.84\times {{10}^{-5}}\sec \\\ \end{aligned}$

Note: We can calculate force by using electric fields. Use the motion of a uniform electric field. There are three types of kinematic equations. Depends on problem we have to use type of kinematic equation: $s=ut+\dfrac{a{{t}^{2}}}{2}$, ${{v}^{2}}={{u}^{2}}+2as$ and $v=u+at$. We know that, if a particle moving initially with velocity ‘u’ is accelerated and move with constant acceleration ‘a’ then its velocity ‘v’ at instant ‘t’ is given by$s=ut+\dfrac{a{{t}^{2}}}{2}$. The displacement ‘s’ at instant of a particle moving with initial velocity ‘u’ and constant acceleration ‘a’ is given by${{v}^{2}}={{u}^{2}}+2as$. The final velocity, initial velocity u constant acceleration’ and displacement’ are related as$v=u+at$. If you know the value of the electric field and type of charge then you can calculate the value of voltage and force.