# Can an electric field be there without a charged partical?

How does a photon have an electric field around itself without any charge particle inside it.

• "Photon is the electric field". More exactly, photon is the gauge boson mediating electromagnetic interaction. Nov 27, 2015 at 9:18

At the risk of phrasing this rather loosely, a charge is needed to create an electric field. If you draw field lines representing your electric field then they must start and end on a charge. More precisely, the divergence of the electric field must be zero unless a charge is present.

However in an infinite plane electromagnetic wave the field lines do not begin or end anywhere, so no charge is required for the field to exist.

This is of course a somewhat artificial model since no electromagnetic wave can be infinite. In practice charge is needed to create the wave, and charge is needed to absorb the wave. The field lines for the wave will begin on the charge that creates it and end on the charge that absorbs it.

Let's start one step behind. We need to look at electric but also at magnetic fields, we can't just explain with one. As you wanted, we assume there is no charge "making" the field. If there is no charge, the charge density is (obviously) 0. Then the current density is 0, too.

$\rho(\vec{r},t)=\vec{j}(\vec{r},t)=0$.

Now, as usual in electromagnetism, lets look at the maxwell equations for this problem:

$\vec\nabla\cdot\vec{E} = 0$

$\vec\nabla \times \vec{B} - \frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}=0$

$\vec\nabla\cdot \vec{B} = 0$

$\vec\nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t}=0$.

I wont solve them here (look the calculus up in your em-book of favour) but special solutions to these eq. are (of course polarisation plays another role here, I'll assume the simplest case):

$\vec{E}(\vec{r},t)=E_0\vec{e}_x\exp{i(kz-\omega t)}$

$\vec{B}(\vec{r},t)=\frac{1}{c}E_0\vec{e}_y\exp{i(kz-\omega t)}$.

So, to answer your question: Yes, there can be an electric field without a charged particle, it directly follows from the maxwell equations. Then you were talking about "photons". Until know in the calculus there were no photons. It may be problematic to introduce it this way (better look up in a book on quantum mechanics) but let me try.

When "seeing" light, we basically talk of waves. Although, when measuring vveerryy low fields we notice, that we actually dont observe a continuus field but particles, photons. And I guess now it should be clear: The em waves are the light, but as the photons are the light, the photons are the em fields (for a shorter "explanation", look at Mikael Kuismas comment).