Electric Potential

The Electric Field \(\vec{E}\) is a vector quantity, which makes it computationally expensive to handle. To simplify the analysis of electrostatics, we introduce a scalar field: the Electric Potential (\(V\)).

The useful idea is that \(V\) stores the same electrostatic information in a scalar form. Once we know the potential at two points, we can compare them directly: moving a charge \(Q\) from \(a\) to \(b\) involves the potential difference \(\Delta V = V_b - V_a\). The electric field does work \(W_{\text{field}} = Q(V_a - V_b)\), so in many problems we can avoid following the full vector field along the path and just compare the endpoint potentials.

1. Definition and The Curl

The definition of potential relies entirely on a specific property of the electrostatic field. Because the field is conservative, its curl is zero everywhere.

\nabla \times \vec{E} = 0

A vector field with zero curl can always be written as the gradient of a scalar. This allows us to define \(V\) such that:

\vec{E} = - \nabla V

To find the potential at a specific point \(\vec{r}\), we integrate the field from a reference point \(\mathcal{O}\):

V(\vec{r}) = - \int_{\mathcal{O}}^{\vec{r}} \vec{E} \cdot d\vec{l}

What is \(\mathcal{O}\)?
The symbol \(\mathcal{O}\) represents the standard reference point where we agree that \(V=0\). For localized charge distributions (like a single atom or a charged ball), we typically place this reference point at infinity (\(\infty\)).

Deep Dive: Derivation & The Curl Condition

The simulator below visualizes \(V\) as a color map. Red means positive potential, blue means negative potential, and black means the potential is close to zero. Brighter color means a larger magnitude \(|V|\). Place charges, move the pointer over the canvas to read \(V\), and try the listed experiments to see how equipotentials change for single charges, dipoles, and multiple charges.

2. Potential of a Localized Charge Distribution

Typically, we know where the charge is (we know \(\rho\)), and we want to find \(\vec{E}\). We could use Coulomb's Law, but vector integrals are difficult.

The strategy is to find \(V\) first. Poisson's equation (\(\nabla^2 V = -\rho/\epsilon_0\)) relates the two, but it is a differential equation. To solve for \(V\) directly from \(\rho\), we essentially "invert" Poisson's equation.

By summing the potential contributions of infinitesimal charge elements \(dq = \rho d\tau'\), we arrive at the superposition integral:

V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|} d\tau'

This scalar integral is generally much easier to compute than the vector integral required for Coulomb's Law. Once we have \(V\), we simply take the gradient (\(-\nabla V\)) to get the field.

Deep Dive: Calculations & Examples

In the next simulator, the left canvas is where you draw a charge distribution and the right canvas shows the resulting potential. The same color rule applies: red is positive \(V\), blue is negative \(V\), black is near zero, and brighter color means a larger \(|V|\). Use it to compare a few separated charges with a smoother continuous distribution.

3. Boundary Conditions

Why do we study boundary conditions? In many problems, we don't know the charge distribution \(\rho\) everywhere. Instead, we have to solve Laplace's Equation (\(\nabla^2 V = 0\)) in different regions and "stitch" the solutions together at the boundaries.

When crossing a surface charge density \(\sigma\), the electric field \(\vec{E}\) suffers a discontinuity. However, the potential \(V\) remains continuous.

V_{\text{above}} = V_{\text{below}} $$ $$ \frac{\partial V_{\text{above}}}{\partial n} - \frac{\partial V_{\text{below}}}{\partial n} = -\frac{\sigma}{\epsilon_0}

This means the potential function does not "jump" at boundaries, but its slope (the normal derivative) changes sharply, corresponding to the discontinuity in the electric field.

Deep Dive: Boundary Conditions