Is there a difference between Electric and Electrostatic Field?

Question

Is there a difference between Electric and Electrostatic Field? All I know is that they both represented with same law suppose we have a Charge placed at the Origin: $$E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}.$$

Answer 1

Electrostatic refers to the case where the fields are not time dependent. In that case the Maxwell's equations reduce to:

$$\nabla \cdot E =\frac{\rho}{\epsilon_o} \\ \nabla \times E = 0 \implies E=-\nabla \phi \\ \text{then,} \nabla \cdot \nabla \phi = \nabla^2 \phi = -\frac{\rho}{\epsilon_o} $$

The solution to the last equation is:

$$ \phi = \frac{1}{4\pi\epsilon_o} \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r}'|}d^3r' $$

which gives the electrostatic electric field equation you have written for point charges.

But if the case is time dependent then you have the Maxwell equation:

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

and you can no longer define $\phi$ as we did above. In this case you have to work with retarded potentials:

$$\psi(\mathbf{r}, t) = \frac{1}{4\pi\epsilon_o} \int \frac{\rho(\mathbf{r'}, t')}{|\mathbf{r}-\mathbf{r}'|}d^3r'$$

where $\psi = \phi, A_x,A_y,A_z$ and $t'=\frac{|\mathbf{r}-\mathbf{r'}|}{c}$. (Note that when $\psi = \phi$ then $\rho$ is the charge density and when $\psi = A_i$ then $\rho = J_i$, i.e., the $i^{th}$ component of current density). Then the fields are given as:

$$ \mathbf{ E} = -\nabla \phi -\frac{\partial \mathbf{A}}{\partial t}\\ \mathbf{B} = \nabla \times \mathbf{A} $$

These time dependent fields turn out to be very complicated expressions and are named the Jefimenko equations.