Question

On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is:
(A) 0.1V
(B) 8V
(C) 2V
(D) 0.5V

Answer 1

Hint: When a force does work on an object, potential energy can be stored. An object with potential energy has the potential to do work.
Formula Used: The formulae used in the solution are given here.
Potential difference between two points in an electric field: ${V_A} - {V_B} = \dfrac{W}{{{q_0}}}$ where ${V_A}$ and ${V_B}$ are the potential of two points A and B in an electric field, $W$ is the work done in moving a charge ${q_0}$ from point A to point B.

Complete Step by Step Solution: Potential difference between two points in an electric field: ${V_A} - {V_B} = \dfrac{W}{{{q_0}}}$ where ${V_A}$ and ${V_B}$ are the potential of two points A and B in an electric field, $W$ is the work done in moving a charge ${q_0}$ from point A to point B.
Given that, when a charge of 20 coulombs is moved by 2 cm, 2J of work is done.
Thus, $W = 20J$ and ${q_0} = 20C$. We assign these values to find the potential difference.
$\therefore $Potential difference = ${V_A} - {V_B} = \dfrac{2}{{20}}$.
$ \Rightarrow {V_A} - {V_B} = 0.1V$

Hence the potential difference is 0.1V.
The correct answer is Option A.

Note: When charges move in an electric field, something has to do work to get the charge to move. To move a charge $q$, we apply a force to just barely overcome the repulsive force from the other charge.
The change of potential energy stored in $q$ is equal to the work done on $q$ to bring it from A to B. Electric potential difference, also known as voltage, is the external work needed to bring a charge from one location to another location in an electric field. Electric potential difference is the change of potential energy experienced by a test charge that has a value of +1. In any electric field, the force on a positive charge is $F = qE$ where the electric field is $E$.