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: $$E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}.$$

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    $\begingroup$ Semantically? An electric field does not have to be static while in electrostatics it's generally assumed that the electric field does not change or only changes so slowly that its rate of change doesn't matter. That's very different in electrodynamics where electric and magnetic field changes are always coupled. $\endgroup$ – CuriousOne Apr 28 '15 at 5:13

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.

  • $\begingroup$ Very well, thank you , I understand now ,I appreciat it $\endgroup$ – user73368 Apr 28 '15 at 7:05
  • $\begingroup$ @user73368 But do understand that they're still the one field - electrostatics are just a special case of the more general electromagnetism. It's just a tool we can use to understand simple electric interactions and do calculations within the given assumptions. The difference is in the map, not the territory. $\endgroup$ – Luaan Apr 28 '15 at 7:17

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