Potential

Explanation¹

The electric field is a special vector function that always has a curl (rotation) of $\mathbf{0}$. From this characteristic, we introduce a scalar function called electric potential. The potential is denoted as $V$ and has the following relationship with the electric field $\mathbf{E}$.

$$ \mathbf{E} = -\nabla V $$

Therefore, if we know the potential $V$, we can know the electric field $\mathbf{E}$. Since the potential is a scalar function, it’s easier to find the potential than to directly find the vector function, the electric field. The relationship between electric field and potential is similar to that between gravity and potential energy. However, since the unit of electric field is not a force, the potential is not exactly potential energy but rather just potential.

The following result means that the electric field is precisely the gradient of the potential, and knowing the potential allows us to calculate its gradient to find the electric field. Moreover, this concept is applicable not only to electric fields but also to all vector functions with a curl of $\mathbf{0}$.

Derivation

Through the process of proving that $\nabla \times \mathbf{E}=0$, we found out that the line integral of the electric field over a closed path is $0$. In the image above, combining path $1$ and path $2$ forms a closed path going from point $\mathbf{a}$ back to point $\mathbf{a}$. Therefore,

$$ \int _{1} \mathbf{E} \cdot d\mathbf{l} + \int_2 \mathbf{E} \cdot d\mathbf{l} =0 $$

the integral values of path $1$ and path $2$ are the same in magnitude but opposite in sign. If we reverse path $1$ so both paths go from point $\mathbf{a}$ to point $\mathbf{b}$, then they have the same value.

Since the line integral of the electric field is independent of the path, the integral from a reference point $\mathcal{O}$ to location $\mathbf{r}$ always has the same value. Therefore, let’s define the scalar function $V$ as follows.

$$ V(\mathbf{r} ) \equiv - \int _\mathcal{O} ^{\mathbf{r}} \mathbf{E} \cdot d \mathbf{l} $$

Then, by the definition of potential, the following formula holds.

$$ \begin{align*} V(\mathbf{b} )- V( \mathbf{a} ) =&\ -\int _\mathcal{O} ^{\mathbf{b}} \mathbf{E} \cdot d\mathbf{l} +\int_\mathcal{O} ^{\mathbf{a}} \mathbf{E} \cdot d \mathbf{l} \\ =&\ -\int_\mathbf{a} ^\mathbf{b} \mathbf{E} \cdot d\mathbf{l} \end{align*} $$

Fundamental Theorem of Gradients
$$ T(b)-T(a) = \int _{a}^{b} (\nabla T) \cdot d\mathbf{l} $$

Also, by the fundamental theorem of gradients, the following formula is valid.

$$ V( \mathbf{b} ) - V (\mathbf{a} ) = \int_{\mathbf{a}}^{\mathbf{b}}\left( \nabla V \right) \cdot d\mathbf{l} $$

Therefore, since $\displaystyle \int_\mathbf{a} ^ \mathbf{b} \left( \nabla V \right) \cdot d\mathbf{l} = -\int_\mathbf{a} ^\mathbf{b} \mathbf{E} \cdot d\mathbf{l}$,