Animated Solution for Physics - Electrostatics: The electric potential $V$ at any point $x,y,z$ (all in metre) in space is given by $V=4x^2$ volt. The electric field at the point $(1\text{ m},0,2\text{ m})$ is ……… V/m.

The relationship between electric potential and the electric field is one of the most elegant concepts in electrostatics. It bridges the scalar world of potential energy with the vector world of forces. Let's dive into this problem and see how a simple scalar function can reveal the entire electric field in a 3D space!

Imagine you are standing in a 3D coordinate system. You are given a map of the electric potential in this space, described by the function:

Notice something interesting? The potential $V$ only depends on the $x$-coordinate. It doesn't care about where you are on the $y$-axis or the $z$-axis. Physically, this means that if you slice the space with planes parallel to the $yz$-plane, every point on a specific plane will have the exact same potential. These are our equipotential surfaces.

To find the electric field $\mathbf{E}$ from the potential $V$, we use the gradient operator. The electric field is the negative gradient of the potential:

In Cartesian coordinates, this expands to:

This equation tells us that the electric field points in the direction where the potential decreases the fastest.

Let's compute the partial derivatives of our potential function $V=4x^2$.

First, we differentiate with respect to $x$:

Next, we differentiate with respect to $y$ and $z$. Since $4x^2$ acts as a constant when we vary $y$ or $z$, these derivatives are simply zero:

Substituting these back into our master equation, we get the general expression for the electric field at any point in space:

We are asked to find the electric field at a specific point $P(1,0,2)$.

We substitute the $x$-coordinate of our point ($x=1$) into our electric field expression. The $y$ and $z$ coordinates ($0$ and $2$) are irrelevant here because the field only depends on $x$.

And there we have it! The electric field at point $P$ points in the negative $x$-direction with a magnitude of $8\text{ V/m}$.

This result perfectly aligns with our physical intuition. Since the potential $V=4x^2$ increases as you move away from the origin along the $x$-axis, the electric field must point back towards the origin (the negative $x$-direction) to point towards the decreasing potential.