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Coulomb's law |
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| Dec23-04, 08:38 PM | #1 |
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Coulomb's law
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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| Dec23-04, 09:25 PM | #2 |
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[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. |
| Dec23-04, 09:56 PM | #3 |
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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?
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| Dec23-04, 10:21 PM | #4 |
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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.
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| Dec24-04, 12:16 AM | #5 |
Recognitions:
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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 |
| Dec24-04, 09:53 AM | #6 |
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thnx got it
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