The Scalar Potential; Poisson and Laplace
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You must be comfortable with the definitions of the gradient, divergence, and Laplacian operators in Cartesian coordinates, and you should recall that the electrostatic field is conservative, meaning its line integral around any closed loop is zero.
Why this matters
Solving for a scalar potential is often easier than solving for a vector field directly.
In electrostatics, we are often given a distribution of charge and asked to find the electric field. Directly calculating the vector sum of Coulomb forces is cumbersome for complex geometries. By introducing the scalar potential, we reduce the problem to finding a single scalar function. Once this function is known, the electric field follows immediately via differentiation. This lecture establishes the fundamental differential equations governing this potential: Poisson's equation, which links the potential to the source charge, and Laplace's equation, which describes the potential in charge-free regions.
The Idea in Plain Language
Imagine a landscape of hills and valleys. The electric field is like the slope of this landscape; it tells you which way a ball would roll and how steeply. The scalar potential is the height of the landscape at any point. If you know the height everywhere, you can calculate the slope (the field) by seeing how the height changes as you move in any direction. Conversely, if you know the slope, you can reconstruct the height, provided the landscape is smooth and has no "swirls" or vortices. In electrostatics, the field has no swirls; it is purely driven by the presence of charge. Therefore, we can always define a height function (the potential) such that the electric field is simply the negative gradient of this height. The negative sign ensures that the field points "downhill," from high potential to low potential, just as water flows downhill.
From Field to Potential
Since the electrostatic field is irrotational, \(\nabla\times\mathbf{E}=\mathbf{0}\), vector calculus guarantees that \(\mathbf{E}\) can be expressed as the gradient of a scalar function \(\varphi\). We define the relationship as:
The minus sign is a convention that ensures the electric field points in the direction of decreasing potential. For a positive point charge, the potential is positive and decreases with distance, so the gradient points inward (towards the charge), and the negative sign makes the field point outward (away from the charge), which matches physical intuition.
Deriving Poisson's Equation
To find the equation that \(\varphi\) must satisfy, we substitute the potential definition into Gauss's law in differential form. Gauss's law states that the divergence of the electric field is proportional to the charge density \(\rho\):
Substituting \(\mathbf{E} = -\nabla\varphi\) into this equation gives:
Using the vector identity that the divergence of a gradient is the Laplacian (\(\nabla\cdot\nabla\varphi = \nabla^2\varphi\)), we obtain:
Rearranging this yields Poisson's equation:
This is a second-order partial differential equation. It tells us that the "curvature" of the potential at any point is directly proportional to the charge density at that point. Where there is positive charge, the potential is "concave down" (a local maximum-like behavior in the vicinity), and where there is negative charge, it is "concave up."
Laplace's Equation and Source-Free Regions
In many practical problems, we are interested in the electric field in regions where there is no charge, such as the interior of a hollow conductor or the space between capacitor plates (ignoring edge effects). In these regions, \(\rho = 0\). Poisson's equation simplifies to:
This is Laplace's equation. It states that in a charge-free region, the potential has no local maxima or minima. The value of the potential at any point is the average of the potential values in its immediate neighborhood. This property is crucial for solving boundary-value problems, where we know the potential on the boundaries of a region and need to find the potential inside. The solution to Laplace's equation is unique for a given set of boundary conditions, a fact we will explore in the next lecture.
The derivation
Scalar Potential, Poisson and Laplace Equations
Because the electrostatic field is irrotational, \(\nabla\times\mathbf{E}=\mathbf{0}\), it can be written as the gradient of a scalar potential, \(\mathbf{E}=-\nabla\varphi\). Substituting this into the differential form of Gauss's law, \(\nabla\cdot\mathbf{E}=\rho/\varepsilon_0\), yields Poisson's equation \(\nabla^2\varphi=-\rho/\varepsilon_0\), whose source-free limit \(\rho=0\) is Laplace's equation \(\nabla^2\varphi=0\).
The potential \(\varphi\) is defined only up to an arbitrary additive constant. Adding a constant \(C\) to \(\varphi\) does not change the electric field \(\mathbf{E} = -\nabla\varphi\), because the gradient of a constant is zero. Therefore, when solving Poisson's or Laplace's equation, you must specify a reference point (e.g., \(\varphi(\infty)=0\) or \(\varphi(0)=0\)) to obtain a unique numerical solution. Without this boundary condition, the solution is indeterminate by a constant. C
Worked example
Problem. Consider a spherical shell of radius \(R\) with a uniform surface charge density \(\sigma\). Find the electric potential \(\varphi(r)\) everywhere in space, taking \(\varphi(\infty)=0\). Then, find the electric field \(\mathbf{E}(r)\).
- Identify the symmetry. The problem is spherically symmetric, so \(\varphi\) depends only on the radial distance \(r\). The Laplacian in spherical coordinates for a function depending only on \(r\) is \(\nabla^2\varphi = \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\varphi}{dr}\right)\).
- Region 1: \(r > R\). Here \(\rho = 0\), so \(\nabla^2\varphi = 0\). The general solution is \(\varphi(r) = \frac{A}{r} + B\). Since \(\varphi(\infty)=0\), we must have \(B=0\). Thus, \(\varphi(r) = \frac{A}{r}\).
- Region 2: \(r < R\). Here \(\rho = 0\) as well (the charge is only on the surface). So \(\nabla^2\varphi = 0\). The general solution is again \(\varphi(r) = \frac{C}{r} + D\). However, the potential must be finite at \(r=0\). If \(C \neq 0\), \(\varphi\) would diverge at the origin. Therefore, \(C=0\), and \(\varphi(r) = D\), a constant.
- Apply boundary conditions. The potential must be continuous at \(r=R\). So, \(\varphi(R^-) = \varphi(R^+)\). This gives \(D = \frac{A}{R}\), so \(A = DR\). The potential is \(\varphi(r) = D\) for \(r < R\) and \(\varphi(r) = \frac{DR}{r}\) for \(r > R\).
- Determine the constant \(D\). Use Gauss's law in integral form for a Gaussian sphere of radius \(r > R\). The total charge is \(Q = 4\pi R^2 \sigma\). The flux is \(\oint \mathbf{E}\cdot d\mathbf{A} = E(r) 4\pi r^2 = \frac{Q}{\varepsilon_0}\). So, \(E(r) = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}\). Since \(\mathbf{E} = -\nabla\varphi = -\frac{d\varphi}{dr}\hat{\mathbf{r}}\), we have \(-\frac{d}{dr}\left(\frac{DR}{r}\right) = \frac{DR}{r^2}\). Equating this to the field from Gauss's law: \(\frac{DR}{r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}\). Thus, \(D = \frac{\sigma R}{\varepsilon_0}\).
- Final potential: \(\varphi(r) = \frac{\sigma R}{\varepsilon_0}\) for \(r < R\), and \(\varphi(r) = \frac{\sigma R^2}{\varepsilon_0 r}\) for \(r > R\).
- Electric field: For \(r < R\), \(\mathbf{E} = -\nabla(\text{constant}) = \mathbf{0}\). For \(r > R\), \(\mathbf{E} = -\frac{d}{dr}\left(\frac{\sigma R^2}{\varepsilon_0 r}\right)\hat{\mathbf{r}} = \frac{\sigma R^2}{\varepsilon_0 r^2}\hat{\mathbf{r}}\).
Units check. \(\sigma\) has units \(\mathrm{C/m^2}\). \(R\) has units \(\mathrm{m}\). \(\varepsilon_0\) has units \(\mathrm{C/(V\cdot m)}\). So \(\frac{\sigma R}{\varepsilon_0}\) has units \(\frac{(\mathrm{C/m^2})\mathrm{m}}{\mathrm{C/(V\cdot m)}} = \frac{\mathrm{C/m}}{\mathrm{C/(V\cdot m)}} = \mathrm{V}\). This is correct for potential. For the field, \(\frac{\sigma R^2}{\varepsilon_0 r^2}\) has units \(\frac{(\mathrm{C/m^2})\mathrm{m^2}}{(\mathrm{C/(V\cdot m)})\mathrm{m^2}} = \frac{\mathrm{C}}{\mathrm{C/(V\cdot m)}} = \mathrm{V/m}\). This is correct for electric field.
Three quick tests
- If \(\nabla^2\varphi = 0\) in a region, what can you say about the charge density in that region?
Answer
The charge density \(\rho\) is zero in that region. - Why is the electric field zero inside a uniformly charged spherical shell?
Answer
By symmetry, the potential is constant inside the shell. Since \(\mathbf{E} = -\nabla\varphi\), the gradient of a constant is zero, so the field is zero. - What is the physical significance of the minus sign in \(\mathbf{E} = -\nabla\varphi\)?
Answer
It ensures that the electric field points in the direction of decreasing potential, which is the direction a positive test charge would naturally move.
Problem set for PU-102 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.