How does permittivity in Coulomb's law work?

In summary, the electric field magnitude is determined by the permittivity and susceptibilty of the medium between the charges.
  • #1
Efeguleroglu
24
2
Coulomb's Law $$ \vec{F} = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r^2} \hat{r} $$
$$ \vec{E} = \frac{1}{4 \pi \epsilon} \frac{Q}{r^2} \hat{r} $$
Let's say we want to find electric field with a distance r from charge Q. How does permittivity effects the magnitude? Will I choose the permittivity at point r or should I take into account all permittivity values between positions 0 and r?

In my opinion I should not only consider permittivity at point r but permittivities between 0 and r are effective as well. Because if we use q1 and q2 and calculate forces acting on each of them, we violate Newton's third law also it is not applicable in coulomb's law. But I don't know how to calculate electric field and thus forces acting on them.

If permittivity values between 0 and r are effective on the magnitude of electric field at r, then we should be able to create a electric field isolator.

I am really confused. I hope someone can help me. Thanks in advance.
 
Physics news on Phys.org
  • #2
Efeguleroglu said:
Will I choose the permittivity at point r or should I take into account all permittivity values between positions 0 and r?
Hi,
My guess is that you are still in an introduction phase. Correct me if I am wrong.
[edit: mixup, see post #2]
And ##\varepsilon## is used as the symbol for susceptibility, not permittivity (that is ##\mu##).

In general we start the study by considering electric fields in vacuum. And the proper way would be to use ##\ \varepsilon_0\ ## to designate the susceptibility.

To answer your (very good) question: not only all permittivity values between 0 and r, but in the entire space! Because the presence of stuff with a different ##\varepsilon## (such as dielectrics) influence the field.

So: for now assume ##\varepsilon = \varepsilon_0\ ## everywhere (except in conductors) until it's clearly stated otherwise.

And:
Efeguleroglu said:
we should be able to create a electric field isolator.
We can: it's called a Faraday cage and it shields what is inside from electric fields

##\ ##
 
Last edited:
  • Like
Likes PeroK
  • #3
My understanding is that μ is the symbol for magnetic permeability, ε is used for electric permittivity and χ for susceptibility.
 
  • Like
Likes BvU
  • #4
Well corrected ! Thanks.
I made a mess of it ! (cause: never use the words, always the symbols o:))
Tried to fix things a little.
 
  • #5
In a capacitor, for this formula
$$ C = \epsilon \frac{A}{d}$$
dielectric constant is calculated using
$$ \epsilon_{eff} = \frac{\int\epsilon dV}{V} $$
or in 2D
$$ \epsilon_{eff} = \frac{\int\epsilon dA}{A} $$
I know capacitors are full of approximations but there is this formula and I don't know how to use it for a more fundamental level.
Maybe this is true:
Suppose we have point charges q1 and q2 with a distance L between them. Then,
$$\epsilon_{eff} = \frac{\int_0^L \epsilon dx}{L}$$
The magnitude of electric force acting on any of them is
$$F= \frac{1}{4 \pi \epsilon_{eff}} \frac{q_1 q_2}{L^2}$$
I am not really sure what I am doing. I'm just assuming arithmetic mean will work in this way. Is this true?
 
  • #6
BvU said:
My guess is that you are still in an introduction phase. Correct me if I am wrong.
##\ ##
Yes, I am just a poor freshman.
 

Related to How does permittivity in Coulomb's law work?

1. How is permittivity defined in Coulomb's law?

In Coulomb's law, permittivity is defined as the measure of a material's ability to store an electric field. It is denoted by the symbol ε and is measured in units of farads per meter (F/m).

2. What is the role of permittivity in Coulomb's law?

Permittivity plays a crucial role in Coulomb's law as it determines the strength of the electric force between two charged particles. It is a constant that reflects the properties of the medium between the charged particles.

3. How does permittivity affect the electric force between two charged particles?

The higher the permittivity of the medium, the weaker the electric force between two charged particles will be. This is because a higher permittivity means the medium can store more electric field, reducing the strength of the electric force.

4. What factors can influence the permittivity in Coulomb's law?

The permittivity of a material can be influenced by factors such as temperature, pressure, and the presence of other charged particles. It can also vary depending on the frequency of the electric field.

5. How is permittivity related to the dielectric constant in Coulomb's law?

The dielectric constant of a material is equal to its permittivity divided by the permittivity of a vacuum. It is a measure of how much a material can reduce the electric field between two charged particles. A higher dielectric constant means a higher permittivity and a weaker electric force between the particles.

Similar threads

Replies
3
Views
777
  • Electromagnetism
Replies
10
Views
1K
  • Electromagnetism
Replies
11
Views
1K
Replies
4
Views
539
Replies
3
Views
756
Replies
2
Views
678
Replies
25
Views
1K
Replies
8
Views
777
Replies
0
Views
704
  • Electromagnetism
Replies
4
Views
1K
Back
Top