How does permittivity in Coulomb's law work?

  • #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.
 

Answers and Replies

  • #2
BvU
Science Advisor
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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:
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:
  • #3
tech99
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My understanding is that μ is the symbol for magnetic permeability, ε is used for electric permittivity and χ for susceptibility.
 
  • #4
BvU
Science Advisor
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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
Efeguleroglu
24
2
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
Efeguleroglu
24
2
My guess is that you are still in an introduction phase. Correct me if I am wrong.
##\ ##
Yes, I am just a poor freshman.
 

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