Calculating Charge and Force of Attraction for Two Charged Coins on a Lab Table

  • Thread starter Thread starter Blink691
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the charge of two identical charged coins placed 1.35 meters apart, each experiencing a force of 2.0 N due to Coulomb's law. The correct formula to use is F = k * q^2 / r^2, where F is the force, k is Coulomb's constant, and r is the distance between the charges. The initial misunderstanding involved incorrectly using the force value as a charge value. The force between like charges is classified as repulsion, not attraction, as both coins possess identical charges.

PREREQUISITES
  • Coulomb's law and its application
  • Understanding of electric charge and forces
  • Knowledge of Coulomb's constant (k)
  • Basic algebra for rearranging equations
NEXT STEPS
  • Learn how to apply Coulomb's law in various scenarios
  • Research the value and significance of Coulomb's constant (k)
  • Explore the concepts of electric charge types: positive and negative
  • Study the differences between attractive and repulsive forces in electrostatics
USEFUL FOR

Students in physics, educators teaching electrostatics, and anyone interested in understanding the principles of electric forces and charges.

Blink691
Messages
17
Reaction score
0
I did this problem, yet I found out that its wrong because I didn't plug the correct force in. I have no idea what to do now though. If someone can show me what I did wrong and explain it, I would greatly appreciate that. Thank you!

Two coins with identical charges are placed on a lab table 1.35 m apart.

A.If each coin experiences a force of 2.0 N because of the presence of the other coin, calculate the charge on each coin.

FE=KE (q1q2/r^2)
R^2 = 1.35m
q1q2 = 2.0N
FE=KE (2.0N*2.0N)/(1.35^2m)
FE=2.194 or 2.195 A

B.Would the force be classified as a force of attraction or repulsion? Explain your answer.

Force of attraction because it represents a weight.
 
Physics news on Phys.org
This question requires a direct application of Coloumb's law (and the assumption that the two coins are like point charges).

Why have you used q1q2 as 2.0N*2.0N? 2 N is the force which each coin experiences, not the charge.

From the information given in the question, you know the values of F , r and that the fact that the two charges are identical. All you need to find is the charge.

Finally, why would the weight decide if the force is attractive or repulsive?
You should know that it is the nature of the charges(ie, like or unlike).
 
Blink691 said:
A.If each coin experiences a force of 2.0 N because of the presence of the other coin, calculate the charge on each coin.

FE=KE (q1q2/r^2)
R^2 = 1.35m
q1q2 = 2.0N
FE=KE (2.0N*2.0N)/(1.35^2m)
FE=2.194 or 2.195 A
(1) 2.0 N is the force, not the charge! The charge is what you are trying to find.
(2) What happened to the constant Ke? (Look it up.)
Start with Coulomb's law:
F = k q^2/r^2
Rearrange that equation to solve for q in terms of F, k, and r (which are all given or known).

B.Would the force be classified as a force of attraction or repulsion? Explain your answer.

Force of attraction because it represents a weight.
This force has nothing to do with weight (weight is due to mass, not charge). Do like charges attract or repel?
 

Similar threads

Replies
10
Views
8K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
5
Views
5K
Replies
3
Views
2K