Question

An electron having charge '$e$' and mass '$m$' is moving in a uniform electric field $E$(neglect gravity). Its magnitude of acceleration will be
A. $\dfrac{{{e^2}}}{m}$
B. $\dfrac{{{E^2}e}}{m}$
C. $\dfrac{{eE}}{m}$
D. $\dfrac{{mE}}{e}$

Answer 1

Hint: Intensity of electric field due to a charge configuration at a point is defined as the force acting on a unit positive charge at that point. Hence, we have to find the answer from the relation between the electrical field and electrical force.

Complete step by step answer:
Intensity of electric field due to a charge configuration at a point is defined as the force acting on a unit positive charge at that point. Electrical force is defined as the natural phenomenon by which a charge exerts force on another charge. The relation between electrical field and force is given by.
$E = \dfrac{F}{q}$
The variables are defined as:
$E = $ Electrical field
$F = $ Electrical force
$q = $ charge
From Newton’s Law we get to know,
$F = ma$
where, $F = $ force, $m = $ mass and $a = $ acceleration

In the given question the charge $q$ of the electron is given as $e$ and mass of electron $m$.Hence, we get,
$F = ma$
According to question,
$q = e$
Substituting the value of $F$ in the relation between electrical field and electrical force,
$E = \dfrac{F}{q} = \dfrac{{ma}}{e}$
By cross-multiplication we get,
$a = \dfrac{{eE}}{m}$
The acceleration of the electron of charge $e$ and mass $m$ is $a = \dfrac{{eE}}{m}$.

Therefore, the correct option is C.

Note: It must be noted that the electrical force depends upon the factors (i) It is directed along a line joining the two particles and is inversely proportional to the square of the separation distance between them. (ii) It is proportional to the product of the magnitudes of the charges of the two particles. (iii) It is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.