Question

A charge is moving from a point A to point B. The work done to move unit charge during this process is called:
A. Potential at A
B. Potential at B
C. Potential difference between A and B
D. Current from A to B

Answer 1

Hint: Since a charge is moved from A to B, it is advised to think on relative terms, when it comes to the potential that the particle posses at A and at B. Remember that work done can be quantized based on this relative potentials relationship. Another approach would be to think of the batteries that we use in circuits. Batteries have a specified voltage. Think of what this voltage entails, given that the batteries have positive and negative terminals.

Formula used:
Work done by a field in moving a charge from one point to another:
$W \propto \Delta PE = q \Delta V \Rightarrow \Delta V = \dfrac{W}{q}$, where $\delta PE$ is the change in potential energy, $\Delta V$ is the potential difference, q is the charge of the particle.

Complete step by step answer:
A good approach to understanding the problem is to gather what happens when a charge is moved from one point to the other, and what the distinction between potential and potential difference is.
Now, consider you have a field of influence, say an electric field E. Suppose you have a test charge (a charged particle) that is currently outside the field of influence at point P. Let’s call this the reference point. Then, to bring the test charge into the field of influence of the electric field from outside the field of influence requires some work to be done, which is restored as a potential of that charged particle. This potential is called electric potential.
Now, you have your test charge in the field of influence of the electric field and it carries some electrical potential. Now, if you were to move your test charge from the point it is at to another point that is within the field of influence of the electric field, then it has a different potential at the new point. Thus, there is a difference in potential in the same particle when moved from one point to another within the influencing field. This is called the potential difference.
To an extent, one can say that the potential is an absolute value whereas the potential difference is a relative value.
Now, electric potential is defined as potential energy per unit charge, i.e.,
$V = \dfrac{PE}{q}$
So, the potential of the charged particle at A: $V_A = \dfrac{PE_A}{q}$ and,
the potential of the charged particle at B: $V_B = \dfrac{PE_B}{q}$ and,
Whereas, the potential difference is defined as
$\Delta V = V_A – V_B = \dfrac{\Delta PE}{q} = \dfrac{PE_A -PE_B}{q}$
This difference in potential is what is of interest to us.
We know that the electrostatic or Coulomb force is a conservative force, which means that work done by the force or the field, on the charged particle, will be independent of the path taken. The work done by a conservative force is the negative of the change in potential energy, i.e., $W= -\Delta PE$. This means that if work done to accelerate a positive charge from rest is positive, it results from a loss in PE, or a negative $\Delta PE$. And subsequently this PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.
Therefore, we have $W \propto \Delta PE = q \Delta V \Rightarrow \Delta V = \dfrac{W}{q}$.
Thus, from the above relation we conclude that the potential difference $\Delta V$ is the amount of work done in moving a unit charge from A to B.

So, the correct answer is “Option C”.

Note:
It is important to note that voltage is the common name for potential difference. For example, every battery has two terminals and its voltage is the potential difference between them. Voltage is this energy stored per unit charge.
Remember that potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. For example, it is much more common to use the concept of voltage (potential energy) than to deal with the Coulomb force directly.