How does permittivity in Coulomb's law work?

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Coulomb's Law incorporates permittivity, represented as ε, which affects the electric field's magnitude. When calculating the electric field at a distance r from a charge Q, it is important to consider the permittivity values throughout the entire space, not just at point r. The discussion highlights the confusion surrounding the application of permittivity in different contexts, such as capacitors and dielectrics. It is suggested that an effective permittivity can be calculated using integrals over the relevant volume or area. Overall, understanding the role of permittivity is crucial for accurately determining electric fields and forces in electrostatics.
Efeguleroglu
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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.
 
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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:
My understanding is that μ is the symbol for magnetic permeability, ε is used for electric permittivity and χ for susceptibility.
 
Well corrected ! Thanks.
I made a mess of it ! (cause: never use the words, always the symbols o:))
Tried to fix things a little.
 
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?
 
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.
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.

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