
#1
Dec2304, 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 



#2
Dec2304, 09:25 PM

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. 



#3
Dec2304, 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?




#4
Dec2304, 10:21 PM

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.




#5
Dec2404, 12:16 AM

HW Helper
Thanks
P: 9,816

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 



#6
Dec2404, 09:53 AM

P: 501

thnx got it



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