# Is really the electric field felt in every thing in space?

I was wondering if the action/force of the electric field is really felt everywhere. I know it does reduce as you get further, but my thoughts concerned more about materials.
So, I know that the electric field is strictly related to Coulomb's Law, and Coulomb's Law is different between materials. So my questions are,

• Does a net elctric field (generated by two oppositely charged object) is felt in every material in space? (In the picture, bottles,wood,copper,air,etc..)
• If it does, does it change its intensity based on materials? How?

I made a picture to try to help you understand my strange question, I drew the electric field reversed, because I like to imagine electrons moving. If there are problem with the question, please let me know in a comment. I will try immediately to fix them, editing the question

• Your question is vaguely touching upon the tribolelectric effect but I'm sure that's not what you want to ask... – Kunal Pawar Apr 20 '17 at 10:13
• I don't know triboeletric effect very much but reading on wikipedia, it seems that there is friction involved. This is not the case – Gabriele Scarlatti Apr 20 '17 at 10:46
• What do you mean when you say material. Do you mean objects or media? – Kunal Pawar Apr 20 '17 at 12:25
• What's the difference? I can have object of the same medium or different medium, I'm saying different like air, glass , wood , copper – Gabriele Scarlatti Apr 20 '17 at 17:48
• There is a difference between the words medium and material. Medium would mean the stuff that separates two objects. Material is rather vague, and from the way you've drawn stuff in your question, one could easily be mistaken that you're talking about objects. – Kunal Pawar Apr 21 '17 at 2:48

If I understood the question correctly, yes.

Since every material, at least at the scale you seem to be interested, is made up of atoms, which is a collection of (balanced) positive and negative charges, it will "feel" the effects of an electric field, since the charges that constitute the material will "feel" it.

Of course different materials (e.g. different types of atoms and arrangements​) respond differently to the same electric fields. A major distinction is to be made between conductors and insulators (or dielectrics). You can find information between the behaviour of these categories of materials in virtually any resource covering electromagnetism.

• Thank you for your answer! I'm wondering, so is the intensity of the electric field the same? I understand different materials can respond differently to an electric field, but is it the value of the field dependent only on distance? – Gabriele Scarlatti Apr 20 '17 at 17:44
• Which electric field? The one inside an object which is a put in an external electric field? Or are you talking about the effects on the electric field sorrounding an object that is charged/put in an external field? As an example the field inside a conductor is zero no matter what, while right on the outside depends on the surface density which itself depends on the shape of the object and on the intensity of the external fields – pp.ch.te Apr 20 '17 at 18:24

Look up something called dielectric constant. It's a measure of how the electric field due to a charge is affected when the charge is placed in another medium. The reference, of course, is made with respect to vacuum.

Vacuum has a dielectric constant $K$=1. While other material media have $K>1$.

Basically the term $K$ occurs in the denominator of the expression for electric field or force and is useful in determining the relative electric permittivity of a medium. The relative electric permittivity $\epsilon_r$, electric permittivity of vacuum $\epsilon_0$ and the dielectric constant $K$ are related by the expression

$\epsilon_r=K\epsilon_0$.

The relative permittivity is a measure of how the electric field/force is affected by the presence of a medium other than vacuum.

For vacuum, any expression for the electric field will be something like $\frac{1}{4π\epsilon_0} \frac{q}{r²}$

For a medium other than vacuum, the expression becomes $\frac{1}{4π\epsilon_r} \frac{q}{r²}$ or
$\frac{1}{4πK\epsilon_0} \frac{q}{r²}$

An example will help. When we say that $K$ is 80 for water, we mean that the electric field or force would be $\frac{1}{80}$ of its value in vacuum.

• That is only partially true, in the sense that is valid only for liquid and linear dielectric. See feynmanlectures.caltech.edu/II_10.html paragraph 10.5 – pp.ch.te Apr 21 '17 at 5:58
• @chiappette Yes I'm well aware of the fact that this holds on for linear isotropic dielectrics. But I think the answer covers what the OP sought to know. – Kunal Pawar Apr 21 '17 at 11:41