Calculating Force Using Coulomb's Law

In summary, the equation states that the force (F) between two objects is proportional to the charges (Q) of the two objects and to the distance between them (r).
  • #1
DB
501
0
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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  • #2
[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.
 
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  • #3
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
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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  • #5
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
 
  • #6
thnx got it
 

1. What is Coulomb's Law?

Coulomb's Law is a physics principle that describes the electrostatic force between two charged objects. It states that the force between two charges is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.

2. How do I calculate the force between two charged objects using Coulomb's Law?

To calculate the force between two charges, you will need to know the magnitude of the charges (q1 and q2) and the distance between them (r). Then, you can use the equation F = k * (q1 * q2)/r^2, where k is the Coulomb's constant (8.99 x 10^9 N⋅m^2/C^2).

3. What is the unit of force in Coulomb's Law?

The unit of force in Coulomb's Law is Newtons (N). This is the SI unit for force and is equivalent to kg⋅m/s^2.

4. Can Coulomb's Law be applied to point charges only?

No, Coulomb's Law can be applied to any charged objects, regardless of their size or shape. However, it is most accurate for point charges, which are small objects with all of their charge concentrated at a single point.

5. How does the distance between two charges affect the force according to Coulomb's Law?

According to Coulomb's Law, the force between two charges is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force decreases. Conversely, as the distance decreases, the force increases.

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