Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

What will be the amount of work done in moving a charged particle of charge $q$ through a potential difference $V$.
A) $\dfrac{q}{V}$
B) $\dfrac{V}{q}$
C) $\dfrac{1}{{qV}}$
D) $qV$

seo-qna
Last updated date: 03rd May 2024
Total views: 337.2k
Views today: 4.37k
Answer
VerifiedVerified
337.2k+ views
Hint: We know the definition of the potential difference. The potential difference is defined as the work done to move a unit charge from one point to another. We can reverse the definition of the potential difference, we can say that the work done to move a unit charge from one point to another is defined as the difference in the electric potential between initial point electric potential and last point electric potential. The electric field is conservative, therefore if we move a charge in the field, opposite to the direction normally the electron moves freely, then we are doing some work and this work is stored as the potential energy.

Complete step by step solution:
By definition of the potential difference express the relation between work and potential difference:
$\therefore V = \dfrac{W}{q}$
Now keep the work done $W$ on one side and the remaining another side
$\therefore W = Vq$
Therefore to move the charge $q$ from one point to another the work done will be $Vq$ .
We can also write it, when the charge is moved from point A to point B, as $W = \left( {{V_B} - {V_A}} \right)q$ , where ${V_A}$ is the electric potential at point A and ${V_B}$ is the electric potential at point B.

Hence the correct option is option D.

Note: When something has potential it means that it has the ability to work. And the work is done by the electric field. A positive charge placed in an electric field will tend to move in the direction of the field. The work done by the electric field, in this case, will be positive as the field and the displacement vectors are in the same direction and the angle between the two vectors will be zero. Otherwise, it will be negative.