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Coulomb's law

by DB
Tags: coulomb
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DB
#1
Dec23-04, 08:38 PM
P: 501
I'm having trouble understanding how to apply the Coulomb's law equation.

F=(1/4 x pi x epsilon) x (q1 x q2/r2)

I know this much:
F(newtons)= {1\4 x pi x [8.854 x 10 - 12 Flow / m] (electrical permittivity)}
x
charge of particle 1 x charge of particle 2/r2 (their distance squared)

Let's say we have two hypothetical particles: 1 with charge 2 C and the other 4 C, they are at a distance of 20 meters apart. I just don't see how to apply this info, would you have to place circle around the 2 particle so that they are at a diameter of 20 apart? If so, then let the radius be 10. Then what?

Thnx to anyone who helps
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Cyrus
#2
Dec23-04, 09:25 PM
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P: 4,780
[tex] F= \frac{|qQ|}{4r^2 \pi \epsilon } [/tex]

Just plug in for your hypothetical case:

[tex] F= \frac{-|2C*4C|}{4 (20)^2 \pi \epsilon } [/tex] (minus sign becuase like repell.)

r is the radial distance between the two points. That just means the distance between them can be thought of as the radius of a circle.
DB
#3
Dec23-04, 09:56 PM
P: 501
wow, it seems so simple know lol, thnx but when you mutiply by electrical permittivity do you just multiply by 8.854 x 10 -12, what happens to the F/m? is it just ignored?

Cyrus
#4
Dec23-04, 10:21 PM
Cyrus's Avatar
P: 4,780
Coulomb's law

The electrical permitivity depends on the type of material you have. The constant you just gave me is for a vaccuum. You can probably find a table in a book somewhere for values in different materials. The permitivity is the value you use for the [tex] \epsilon [/tex]. As for F/m, I dont understand your question im sorry.
ehild
#5
Dec24-04, 12:16 AM
HW Helper
Thanks
P: 10,387
Quote Quote by DB
wow, it seems so simple know lol, thnx but when you mutiply by electrical permittivity do you just multiply by 8.854 x 10 -12, what happens to the F/m? is it just ignored?
Farad/m is the unit of the permittivity, and it is not ignored, but together with the other units - unit of charge, unit of length it makes the unit of force, newton (N). But this happens automatically if you use the SI units.

Lets see what is the dimension of (charge^2)/(permittivity*length^2)?

[tex] \frac {A^2\cdot s^2 }{(F/m) \cdot m^2} =( \frac{As}{m}) (\frac {Q}{F}) = \frac{V \cdot A \cdot s}{m}= \frac{J}{m} = N [/tex]

It has been used that charge = current *time, capacitance = charge/voltage, and power = current* voltage,
work = power *time, work = force * length.

ehild
DB
#6
Dec24-04, 09:53 AM
P: 501
thnx got it


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