When does when elect to use Laplace's equation when dealing with charge distributions. For example, if I had a metallic sphere of radius $R$ and charge $Q$, then

$$\mathbf E = \begin{cases} 0, & \text{for } r < R \\ \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}, & \text{for } r \ge R \end{cases}$$

The potential at all regions of space can be calculated in a straight-forward way then.

For $r \ge R$,

$$V = - \frac{Q}{4 \pi \epsilon_0}\int_{\infty}^r 1/r^2 \ dr$$

And for $r < R$

$$\implies V = 0$$

However, I could've just as easily tried to work it out by saying for $r > R$ and $r<R$:

$$\nabla ^2 V = 0$$

As all charge resides on the surface, and then say that

$$V(r,\theta) = \sum_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l (\cos{\theta})$$

Wouldn't this be an equally valid way to derive the field? If so, when is it better to derive things this way rather than the way I did with Gauss's Law?


2 Answers 2


The first thing to be pointed out is the Gauss's law and Laplace's equation are essentially the same thing (as long as you include the definition of potential as well) since in general for a surface $S$ enclosing a volume $V$ that has some charge distribution $\rho$, $$\iint_S\mathbf E\cdot\text d\mathbf A=\frac{Q_{enc}}{\epsilon_0}=\frac{1}{\epsilon_0}\iiint_V\rho\ \text dV$$ Using the divergence theorem we can change the left hand side of the equation $$\iiint_V\nabla\cdot\mathbf E\ \text dV=\frac{1}{\epsilon_0}\iiint_V\rho\ \text dV$$ And since this is true for any arbitrary volume bound by a surface, it must be that $$\nabla\cdot\mathbf E=\frac{\rho}{\epsilon_0}$$ then using the definition of potential as well as assuming a charge free region we have Laplace's equation $$\nabla^2 V=0$$

However, even though these two equations are saying the same thing and are always valid, it does not mean that they are equally applicable in every situation. Gauss's law is only good at determining the field (and hence the potential) for specific systems with just the right symmetry. These systems are spheres (including point charges), infinite cylinders (including line charges), and infinite sheets (including infinitely thin sheets).

The reason the symmetries are needed for making use of Gauss's law is because they allow us to "pull the electric field out from the integral". In other words, you use the symmetry to make an argument that on your Gaussian surface the field is either

1) constant and perpendicular to certain parts of the surface or

2) parallel to the other parts of the surface.

This allows you to change your integral to $$\iint_S\mathbf E\cdot\text d\mathbf A=E\iint_{S'}\text dA$$ where the integral on the right is just over the sections of the Gaussian surface where the field is perpendicular to our surface. If you can also determine the total charge enclosed by your surface then you can easily determine $\mathbf E$ at all points in space.

The reason we need these symmetries is because the integral is essentially an infinite sum, and just because you know the value of this sum does not mean you know the value of the individual terms. The analogy I like to give is the following: If I told you I have ten numbers that add up to $100$ there would be no way for you to tell me what those ten numbers are. However, if I tell you that these numbers are all equal, then you can easily see that each number is equal to $10$.

Laplace's equation is much more widely applicable. First, because it is an equation dealing with a scalar function $V$ rather than a vector function $\mathbf E$, and second because it is a differential equation which are typically not too bad to solve. Even if there isn't a nice solution you can write down using elementary functions you can always numerically solve the differential equation.

So, in summary, use Gauss's law when symmetry allows you to. Otherwise Laplace's equation (or more generally Poisson's equation) is the way to go.

  • 1
    $\begingroup$ +1, yes they are the same thing in different forms. $\endgroup$
    – user196418
    May 10, 2019 at 20:32

Thought I'd add a little to Aaro Stevens great answer above.

The important two Maxwell's Equations here are:

$$\nabla \cdot \vec{E}=\rho/\epsilon_0$$ $$\nabla \times \vec{E}=0$$

Note the RHS of the second equation is only zero here because we have no time varying electric field.

It can be proven that if the curl of a vector function vanishes, it can be expressed as a gradient, $\vec{E}=-\nabla V$.

We get Poisson's equation by substituting the potential into the first of these equations.


$\rho$ is zero outside of the charge distribution and the Poisson equation becomes the Laplace equation.

Gauss' Law can be used for highly symmetric systems, an infinite line of charge, an infinite plane of charge, a point charge. A less familiar use, it can be used to calculate the flux through highly symmetric regions. For example, if you have a point charge and you want to find the flux through plane.

Poisson's Equation is always applicable including in the absence of useful symmetries and offers some of its own tools for solution. You can often solve the associated differential equations analytically or allowing good approximations via terms from infinite series. There are also various ways of solving them numerically. One being the Method of Relaxation

  • 1
    $\begingroup$ Not to be too pedantic, but Gauss's Law ought to be $\nabla \cdot \mathbf E = \rho/\epsilon_0$ $\endgroup$
    – sangstar
    May 10, 2019 at 19:33
  • $\begingroup$ @sangstar It depends on the convention used $\endgroup$ May 10, 2019 at 20:19
  • $\begingroup$ True! corrected. $\endgroup$
    – R. Romero
    May 10, 2019 at 20:24

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