Animated Solution for Physics - Electrostatics: On moving a charge of $20\text{ C}$ by $2\text{ cm}$, $2\text{ J}$ of work is done, then the potential difference between the points is

Moving the Charge

Imagine you are an external agent tasked with moving a massive $20\text{ C}$ charge through an electric field. You pick up this charge at point $A$ and carry it over to point $B$. The problem tells us that the physical distance between these two points is $2\text{ cm}$.

As you push this charge against the invisible forces of the electric field, you expend energy. The total work you do in this process is exactly $2\text{ J}$. Our goal is to find the electric potential difference between point $A$ and point $B$.

Potential Difference

To solve this, we need to recall the fundamental definition of electric potential difference. What does it actually mean?

Potential difference, denoted as $\Delta V$, is defined as the work done per unit charge to move a test charge between two points. Mathematically, it is expressed as:

This elegant equation tells us that the potential difference is simply the ratio of the total energy expended ($W$) to the magnitude of the charge being moved ($q_0$).

Redundant Information

Before we plug in the numbers, let's address the elephant in the room: the $2\text{ cm}$ distance. Why is it there?

In many physics problems, examiners include extra information to test your conceptual clarity. Because the electrostatic force is a conservative force, the work done in moving a charge between two points is completely independent of the path taken or the physical distance between them. The potential difference depends only on the initial and final states. Therefore, the $2\text{ cm}$ is a classic distractor. We can safely ignore it!

Now, let's substitute our known values into the master equation. We have $W=2\text{ J}$ and $q_0=20\text{ C}$:

Simplifying this fraction, we get:

And there we have it! The potential difference between the two points is exactly $0.1\text{ V}$.