Calculating Vector Force Between Charged Particles

  • Thread starter Minihoudini
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In summary, the problem involves two charged particles, A and B, with a distance of 13.7 mm between them. When particle B moves away to a distance of 17.7 mm, the force exerted on A is requested. Using the equation F= (K q1 q2/r^2), the constant K=8.99 x 10^9, and the known force and distances, we can calculate the product of the charges (q1 q2) and then use it to find the new force at the new distance.
  • #1
Minihoudini
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Homework Statement


A charged particle A exerts a force of 2.62microNewtons to the right on a charged particle B when the particles are 13.7 mm apart. Particle B moves straight away from A to make the distance between them 17.7mm. What vector force does it then exert on A?


Homework Equations


I know its easy, very, but for the love of god I can't figure this out.


The Attempt at a Solution


I know that they are both positive particles, so I am assuming both are +1.60x10^-19c.
Also I believe we use F= (K q1 q2/r^2). I've tried inserting different things into the formula but none of it gives me the answer I am looking for which is 1.57microNewtons
 
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  • #2
Yes, use use F= (K q1 q2/r^2).
No need to assume a charge. Put in the force and distance for the first position, and calculate the value of K q1 q2. This will carry over to the new distance.
Knowing that and the new distance, you can calculate the new force.
 
  • #3
Minihoudini said:
... I am assuming both are +1.60x10^-19c.
I wouldn't make that assumption ... I see nothing in the problem statement that would allow this.
Also I believe we use F= (K q1 q2/r^2).
Yes, that is the equation for this problem.
Here's a question: what must the product (q1 q2) be equal to, so that the force is 2.62 μN when r=13.7mm?

EDIT: Delphi gave even better advice. What is (K q1 q2) when the force is 2.62 μN and r=13.7mm?
 
  • #4
alright, well K=8.99 x 10^9

so the formula should look like this now. r x [square root of F/k] = q1 q2
inputting everything it looks like this
(0.0137m) times square root of (2.62 x 10^-7 N )/(8.99 x 10^9) = 7.39 x 10^-11
equaling that to q1 q2. this is one part I am stuck on, I'll be able to get the rest but I don't know what to do with both unknowns.
 
  • #5
If q1 q2 is bothering you, make this change in the formula: C = q1 q2.
Then you will have only the one unknown!
The important thing is that C is a constant in this problem.
Even better: let C = kq1 q2. Saves wear on the calculator!
 
  • #6
Minihoudini said:
alright, well K=8.99 x 10^9

so the formula should look like this now. r x [square root of F/k] = q1 q2

Uh, wait, that should be
r^2 x F/k = q1 q2​
 
  • #7
there we go, again thanks everyone. that was what was screwing up my answer, I saw another equation like the one I put up for a similar question so I thought I could apply it here. I see now that I can't. again thanks.
 

Related to Calculating Vector Force Between Charged Particles

1. What is Coulomb's force?

Coulomb's force is a fundamental concept in physics that explains the attraction or repulsion between two charged particles. It states that the magnitude of the force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

2. What is the formula for Coulomb's force?

The formula for Coulomb's force is F = k * (q1 * q2) / r^2, where F is the force, k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

3. Is Coulomb's force a vector or scalar quantity?

Coulomb's force is a vector quantity, meaning it has both magnitude and direction. The direction of the force is determined by the charges of the particles (opposite charges attract, like charges repel) and the direction of the distance between them.

4. How does the distance between two charged particles affect Coulomb's force?

The force between two charged particles is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases, and as the distance decreases, the force increases. This relationship is known as the inverse square law.

5. What are the SI units for Coulomb's force?

The SI units for Coulomb's force are Newtons (N). This is because force is measured in Newtons, charge is measured in Coulombs, and the Coulomb constant has units of N*m^2/C^2.

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