Courses for Kids
Free study material
Offline Centres
Store Icon

A particle having mass m and charge q is at rest. On applying a uniform electric field E on it, it starts moving. What is the kinetic energy when it travels a distance x in the direction of force?
A. ${q^2}Ex$
B. $q{E^2}x$
C. $qE{x^2}$
D. $qEx$

Last updated date: 13th Jun 2024
Total views: 403.5k
Views today: 12.03k
403.5k+ views
Hint:- This question can be solved by applying the work-energy theorem to the electric concept. The work-energy theorem states that:
The net work done by the forces on a object translates into the total kinetic energy of the body
$W = KE$.

Complete step-by-step solution:-
Before we solve the problem, we need to understand the fundamental concept of the electric field.
The electric field can be thought of as a sphere of influence around a charge, at which the influence of the charge exists and depends on the distance from the charge, such that when another charge is in the vicinity of this charge, it experiences a force.
The force experienced by this charge is given by Coulomb’s law which states that:
The magnitude of the electrostatic force of attraction or repulsion between any of the two point charges is directly proportional to product of magnitudes of these charges and inversely proportional to square of the distance between them and this force is along the straight line joining them.
$F = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
where ${q_1}\& {q_2}$ are the charges and r is the distance between the charges. (${\varepsilon _o}$ - absolute permittivity).
Even though the electric field might seem like a concept, it is quantifiable too.
The quantity electric field is defined as the force experienced by a unit positive charge when it is placed in the vicinity of a charge $q$ at a distance of r from it.
In the above expression of force, if we substitute the value of the charge ${q_1}$ as $q$ and ${q_2} = + 1C$ or the unit positive charge, the force will be equal to the electric field.
$F = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
$F = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{q\left( 1 \right)}}{{{r^2}}} = E$
In that case, the force on a charge q can be obtained by multiplying the electric field E by the charge q.
$F = qE$
The work done by the force while travelling through the distance $x$ is given by,
$W = F \times x$
$W = qEx$
By work-energy theorem,
$W = KE$
So, the kinetic energy is,
$KE = qEx$
Hence, the correct option is Option D.

Note:- In the expression of the electric force, the value of the constant is,
$K = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
where the value of absolute permittivity, ${\varepsilon _o} = 8 \cdot 85 \times {10^{ - 12}}F{m^{ - 1}}$, which represents the property of the material space in between the charges, to effectively, allow the electric field to pass through them.