Animated Solution for Physics - Electrostatics: An electric charge $10^{-3}\mu \text{C}$ is placed at the origin $(0,0)$ of $xy$-coordinate system. Two points $A$ and $B$ are situated at $(\sqrt{2},\sqrt{2})$ and $(2,0)$ respectively. The potential difference between the points $A$ and $B$ will be

Imagine a coordinate system where a tiny electric charge, $q=10^{-3}\mu \text{C}$, is placed exactly at the origin $(0,0)$. We are given two specific points in this space: point $A$ at $(\sqrt{2},\sqrt{2})$ and point $B$ at $(2,0)$. Our goal is to find the potential difference between these two points.

At first glance, the coordinates of point $A$ might look a bit intimidating with those square roots, but let's not rush. The key to solving this problem lies in understanding what electric potential actually depends on.

The electric potential $V$ at any point in space due to a point charge $q$ is given by the formula:

Here, $r$ is the straight-line distance from the charge to the point in question. Notice that the potential depends only on the distance $r$ and the magnitude of the charge $q$. It does not depend on the direction or the specific $x$ and $y$ coordinates, as long as the radial distance remains the same.

Let's calculate the distance of point $A$ from the origin using the standard distance formula:

Squaring the terms inside the root gives us:

Now, let's look at point $B(2,0)$. Since it lies directly on the x-axis, its distance from the origin is simply its x-coordinate:

Look at that! Both points $A$ and $B$ are exactly $2$ units away from the origin.

Because the distances are identical, the electric potential at both points must be exactly the same.

Therefore, the potential difference between point $A$ and point $B$ is simply zero.

Physical Insight: Points $A$ and $B$ lie on the same equipotential surface—a spherical shell of radius $2$ centered at the origin. Moving any charge along this surface, or between any two points on this surface, requires absolutely zero work by the electric field!