Animated Solution for Physics - Electrostatics: A uniform electric field pointing in positive x-direction exists in a region. Let $A$ be the origin, $B$ be the point on the x-axis at $x=+1\text{ cm}$ and $C$ be the point on the y-axis at $y=+1\text{ cm}$. Then the potentials at the points $A$, $B$ and $C$ satisfy

Title: The River of Electric Potential

Imagine you are standing on the bank of a wide, smoothly flowing river. The water moves steadily in one direction. If you drop a leaf into the water, it naturally drifts downstream. In the world of electrostatics, a uniform electric field is very much like this river, and the concept of electric potential is like the elevation of the riverbed. Water naturally flows from higher elevation to lower elevation. Similarly, an electric field always points from regions of higher electric potential to regions of lower electric potential.

In our problem, we are given a uniform electric field $\vec{E}$ that points strictly in the positive x-direction. We have three points of interest: - Point $A$ is at the origin $(0,0)$. - Point $B$ is on the x-axis at $(1,0)$. - Point $C$ is on the y-axis at $(0,1)$.

Let's visualize this. The electric field lines are horizontal arrows pointing to the right.

What happens if you move perpendicular to the river's flow? You aren't going upstream or downstream, so your elevation doesn't change. In physics, if you move perpendicular to the electric field, the electric potential remains constant. These paths are called equipotential lines (or surfaces in 3D).

Since our electric field is entirely in the x-direction, any vertical line (parallel to the y-axis) is an equipotential line. Points $A$ and $C$ both lie on the y-axis. Therefore, they must have the exact same electric potential!

Now, let's look at point $B$. To get from point $A$ to point $B$, you have to move along the positive x-axis. You are moving exactly in the direction of the electric field—you are going "downstream."

Because the electric field points from higher potential to lower potential, moving in the direction of the field means the potential must be decreasing. Therefore, the potential at point $B$ is strictly less than the potential at point $A$.

Let's check the given options. We know that $V_A$ cannot be less than or greater than $V_C$ because they are equal. This eliminates options (c) and (d). We also know that the potential at $A$ is greater than the potential at $B$, which perfectly matches option (b).

The beauty of this problem lies in its simplicity. You don't need to calculate any exact numbers; you just need to understand the geometric relationship between the electric field and electric potential. Always remember: the electric field points downhill!