Why do we need a second equation for electric field in Maxwell's Equation?

Question

Suppose we are dealing with electrostatics for this question. A physicist carries out experiments with static charges and determines that, the electric field $\vec { E } (\vec { r } )$ is a quantity which behaves as,

$$\nabla .\vec { E } =\frac { \rho }{ { \epsilon }_{ 0 } } $$

Further he notices that, this quantity drops to zero as $\vec { r } \rightarrow \infty$

Is it possible to deduce the field from these conditions alone?

Posed mathematically, the question is that the divergence equation is a first order PDE so by giving enough boundary conditions we should be able to determine the field right? If this was so then why do we need the curl equation? And again if one were to use the curl equation then we would have 3-unknowns and 4 equations, so some of them must be redundant right?

Note:Assume that there is no magnetic field present for the purpose of this question.

Answer 1

Your statement is correct within your stipulation of stationary charges and no time-varying magnetic fields.

Outside of the restricted cases where a) there are no time varying magnetic fields present and b) the electric field is conservative, i.e. is the gradient of a scalar potential, , we need the curl equation

$$\nabla \times \vec E = - \frac {\partial \vec B} {\partial t}$$

to explain the results of additional experiments (starting with Faraday), namely those involving electric fields resulting from electromagnetic induction in a time varying magnetic field.

Answer 2

This is an electrostatic problem. This means, as you mentioned, that $$\nabla \times{\vec E}=0$$ which is equivalent to the existence of an electrostatic potential $\Phi(\vec r)$ so that $$\vec E(\vec r)=-\nabla \Phi(\vec r)$$ Inserting this into $$\nabla \vec { E } =\frac { \rho }{ {\epsilon }_{ 0 } } \tag{I}$$ yields $$\Delta \Phi =-\frac { \rho }{ { \epsilon }_{ 0 } } $$ which is the Poisson equation for the electrostatic potential. This has a unique solution for $\vec E=-\nabla {\Phi}\rightarrow 0$ when $\vec { r } \rightarrow \infty$.

Another point of view is that, mathematically, using the Green's function of the equation (I), which corresponds to Coulomb's law, a solution to equation (I) with the boundary condition $\vec E\rightarrow 0$ when $\vec { r } \rightarrow \infty$ is $$\vec E(\vec r)= \int {\frac{\rho(\vec r') (\vec r - \vec r')d^3r'}{4 \pi \epsilon_0|\vec r - \vec r'|^3}}$$

Note: Equation (I) is also known as the differential form of Gauss's Law. Gauss Law is equivalent to Coulomb's Law. Gauss's law follows from Coulomb's Law and vice versa.

Answer 3

".... the question is that the divergence equation is a first order PDE so by giving enough boundary conditions we should be able to determine the field right?"

Not so. As mentioned in the comments the answer to this question is essentially about the Helmholtz decomposition, but let's actually partly delve into a certain proof of this decomposition that shows very clearly, geometrically and intuitively what the problem is, at least for a large class of vector fields, namely, those that have Fourier transform, as discussed in my answer here.

Imagine a vector field's Fourier decomposition $\vec{F}(\vec{k})$, a function of the plane wave wavevector $\vec{k}$, i.e. we decompose a vector valued function $\vec{f}(\vec{r})$ of position $\vec{r}$ into a superposition of plane wave vector fields of the form $\vec{F}(\vec{k})\,\exp(i\,\vec{k}\cdot\vec{r})$.

Now, how do divergence and curl look in Fourier space? $\nabla\cdot \vec{f}$ has the Fourier transform $\vec{k}\cdot\vec{F}$ and $\nabla\times \vec{f}$ has the transform $\vec{k}\times\vec{F}$; you should be able to prove this fairly straightforwardly.

So now, ask your question in Fourier space terms. It is, "why can we determine the vector $\vec{F}$ from $\vec{k}\cdot\vec{F}$ alone?". It should be very clear that this can't be done; we need to know the components of $\vec{F}$ that are orthogonal to $\vec{k}$ and these can be assigned essentially independently, since the divergence of a vector field everywhere orthogonal to $\vec{k}$ vanishes.

In general, one can assign a smooth scalar field in Fourier space $g(\vec{k})$ and a second smooth vector field $\vec{H}(\vec{k})$ that is everywhere orthogonal to $\vec{k}$, but otherwise arbitrary. As I discuss in this answer here and here, the information $g(\vec{k})$ and $\vec{H}(\vec{k})$ are exactly the information to determine a vector field $\vec{F}$ such that:

$$g(\vec{k}) = \vec{k}\cdot\vec{F}(\vec{k})$$ $$\vec{H}(\vec{k})= \vec{k}\times \vec{F}(\vec{k})$$

So the answer to your question is essentially that the divergence condition tells you only the component of the vector field that is along the wavevector; the solenoidal part, that orthogonal to the wavevector, is missing (it has zero divergence) and can be assigned independently.