Question

An electron moving with the speed $5\times {{10}^{6}}m/s$ is shooted parallel to the electric field of intensity $1\times {{10}^{3}}N/C$. Field is responsible for the retardation of motion of electrons. Now evaluate the distance travelled by the electron before coming to rest for an instant. (Mass of e=$9\times {{10}^{-31}}kg$, charge =$1.6\times {{10}^{-19}}C$ )
A. 7m
B. 0.7mm
C. 7cm
D. 0.7cm

Answer 1

Hint: The electric field is causing a retardation of the motion of the electron. We can find this retardation by equating Newton’s second law to the electric force due field E. The retardation thus found can be substituted in Newton’s equation of motion to find the distance covered before coming to rest ( final velocity is zero).

Formula used:
Expression for electric force,
$F=qE$
Newton’s second law,
$F=ma$
Newton’s equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$

Complete step-by-step answer:
In the question, we are given an electron that is moving at a speed of $5\times {{10}^{6}}m/s$ . And this electron is then shooted parallel to an electric field of intensity $1\times {{10}^{3}}N/C$ . The electron suffers from retardation due to the presence of this field.
We are asked to find the distance covered by the electron before coming to rest for an instant.
We know that, electric force F on a charge q due to the presence of the field E is given by,
$F=qE$ …………………………………. (1)
But from Newton’s second law we have force as the product of mass and acceleration, that is,
$F=ma$ ………………………………. (2)
From the question we know that the electric field is causing the retardation of the electron. So, the ‘a’ here is the retardation of electrons caused by electric force due to electric field E. Therefore, we can now equate RHS of equations (1) and (2).
$\Rightarrow qE=ma$
$\Rightarrow a=\dfrac{qE}{m}$…………………………….. (3)
We are given:
$q=1.6\times {{10}^{-19}}C$
$E=1\times {{10}^{3}}N/C$
$m=9\times {{10}^{-31}}kg$
Now (3) becomes,
$a=\dfrac{1.6\times {{10}^{-19}}\times 1\times {{10}^{3}}}{9\times {{10}^{-31}}}=0.18\times {{10}^{15}}m{{s}^{-2}}$ ……………………….. (4)
We are given the initial velocity u of the electron as,
$u=5\times {{10}^{6}}m/s$ ………………………… (5)
We are asked to find the distance covered by electrons before it comes to rest. So, the final velocity of electron v is given by,
$v=0$ ……………………………… (6)
From Newton’s equations of motion we have,
${{v}^{2}}-{{u}^{2}}=2as$ …………………………. (7)
Where,
v = final velocity
u = initial velocity
a = acceleration
s = distance travelled
Substituting (4), (5) and (6) in (7),
$0-{{\left( 5\times {{10}^{6}} \right)}^{2}}=2\left( 0.18\times {{10}^{15}} \right)s$
Negative sign here indicates the retardation of the electron.
$\Rightarrow s=\dfrac{{{\left( 5\times {{10}^{6}} \right)}^{2}}}{2\times 0.18\times {{10}^{15}}}$
$\Rightarrow s=69.44\times {{10}^{-3}}m=0.069m$
$s=6.9cm\approx 7cm.$
Therefore, the distance travelled by the electron before coming to rest for an instant is 7cm.

So, the correct answer is “Option C”.

Note: First half of the question demands your knowledge on electric force due to the electric field and the second half is solved using Newton’s laws of motion, that is, in the same question we are dealing with two different concepts. In this question you should take care of the powers of 10, failing to do so will lead you to different answers.