# Equipotential Curves

Recall from the Conservative Vector Fields page that we called a vector field $\mathbf{F} = \mathbf{F}(x, y)$ on $\mathbb{R}^2$ conservative if there exists a function $\phi = \phi (x, y)$ such that $\mathbf{F}(x, y) = \nabla \phi (x, y)$. If such a function $\phi$ exists, then we call it a potential of $\mathbf{F}$. The level curves of the function $\phi$ has a special name which we define below.

Definition: If $\mathbf{F} = \mathbf{F}(x, y)$ is a conservative vector field and $\phi = \phi (x, y)$ is a potential of $\mathbf{F}$ then the level curves $\phi (x, y) = C$ are called the Equipotential Curves of $\mathbf{F}$. |

One important property of equipotential curses is that they intersect the field lines of $\mathbf{F}$ at right angles. Let’s look at an example.

Consider the vector field $\mathbf{F}(x, y) = x^2 \vec{i} + y \vec{j}$. It shouldn’t be surprising to see that $\mathbf{F}$ is a conservative vector field. We first note that $\frac{\partial}{\partial y} (x^2) = 0 \frac{\partial}{\partial x} (y)$, so the necessary condition for a vector field to be conservative is satisfied. Furthermore we have that:

(1)

(2)

Thus $\frac{x^3}{3} + m(y) = n(x) + \frac{y^2}{2}$ which implies that $n(x) = \frac{x^3}{3}$ and $m(y) = \frac{y^2}{2}$. Therefore a potential of $\mathbf{F}$ is given by:

(3)

The level curves of this potential are given by $\phi (x, y) = C$, that is $\frac{x^3}{3} + \frac{y^2}{2} = C$. The following image shows the vector field $\mathbf{F}$ alongside some level curves $\phi (x, y) = C$ for different $C$. Notice how the field lines that perpendicular to these level curves as expected.

# Equipotential Surfaces

We will now extend what we saw above to vector fields on $\mathbb{R}^3$.

Definition: If $\mathbf{F} = \mathbf{F}(x, y, z)$ is a conservative vector field and $\phi = \phi (x, y, z)$ is a potential of $\mathbf{F}$ then the level curves $\phi (x, y, z) = C$ are called the Equipotential Surfaces of $\mathbf{F}$. |

We note that if $\mathbf{F} = \mathbf{F}(x, y, z)$ is a conservative vector field and $\phi = \phi (x, y, z)$ is a potential of $\mathbf{F}$ then the level surfaces, $\phi (x, y, z) = C$ are perpendicular to the field lines of $\mathbf{F}$.

For example, consider the vector field $\mathbf{F} (x, y, z) = x\vec{i} + y^2 \vec{j} + z^3 \vec{k}$. It’s not hard to show that this vector field is conservative and that $\phi (x, y, z) = \frac{x^2}{2} + \frac{y^3}{3} + \frac{z^4}{4}$ is a potential of $\mathbf{F}$ as you should verify. The image below shows the vector field $\mathbf{F}$ alongside the level curves $\phi (x, y, z) = 0$, $\phi (x, y, z) = 2$, $\phi (x, y, z) = 10$:

$\phi (x, y, z) = 0$ | |
---|---|

$\phi (x, y, z) = 2$ | |

$\phi (x, y, z) = 10$ |