Calculating Force Using Coulomb's Law

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Discussion Overview

The discussion revolves around the application of Coulomb's law for calculating the force between two charged particles. Participants explore the formula, its components, and how to correctly apply it in hypothetical scenarios, including considerations of electrical permittivity and unit dimensions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses confusion about applying Coulomb's law and visualizing the distance between two charged particles.
  • Another participant provides a formula and suggests substituting values for the hypothetical charges and distance, noting the importance of the radial distance.
  • A participant questions how to handle the electrical permittivity in calculations and its implications for the units involved.
  • Discussion includes clarification that electrical permittivity varies with material and that the constant provided is specific to a vacuum.
  • Further elaboration on the dimensional analysis of the formula is presented, explaining how units combine to yield force in Newtons.

Areas of Agreement / Disagreement

Participants demonstrate a mix of understanding and confusion regarding the application of Coulomb's law and the role of electrical permittivity. There is no clear consensus on the best approach to the problem, and some questions remain unresolved.

Contextual Notes

Participants reference specific constants and units, but there are uncertainties regarding the application of these in different materials and the implications for calculations.

DB
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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? :eek:

Thnx to anyone who helps o:)
 
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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 because 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.
 
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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?
 
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 don't understand your question I am sorry.
 
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DB said:
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
 
thnx got it
 

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