Coulomb's Law and repulsive force

In summary, when the distance between two charged bodies is doubled, the force between them decreases by a factor of 4. This can be seen through the use of the inverse square law, where the force is inversely proportional to the square of the distance between the charges.
  • #1
Soaring Crane
469
0

Homework Statement



Charges q1 and q2 exert repulsive forces of 15 N on each other. What is the repulsive force when their separation is increased so that their final separation is 160% of their initial separation?

a.1.3 N

b.9.8 N

c.8.7 N

d.5.9 N


Homework Equations



F = (k*q_1*q_2)/(r^2)

The Attempt at a Solution



F_2 = [k*q_1*q_2]/[(1.6^2)*(r^2)]

F_2 = F/(1.6^2) = F/2.56

F_2 = 15 N/2.56 = 5.86 N ??




Homework Statement



The force between two very small charged bodies is found to be F. If the distance between them is doubled without altering their charges, the force between them becomes


a.F/2

b.2F

c.F/4

d.4F


Homework Equations



F = (k*q_1*q_2)/(r^2)

The Attempt at a Solution



F_2 = (k*q_1*q_2)/(2r)^2 = (k*q_1*q_2)/(4*r^2)

F_2 = F/4 ?


Thanks.
 
Physics news on Phys.org
  • #2
Yeah I think you've nailed the inverse square concept.
 
  • #3


I would say that Coulomb's Law is an important principle in understanding the forces between charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In the first problem, we are given two charged particles with a repulsive force of 15 N between them. We are asked to find the repulsive force when their separation is increased to 160% of their initial separation. Using Coulomb's Law, we can calculate the new force by dividing the original force by 2.56 (as the separation is increased by 160%, the distance between them is multiplied by 1.6, and the force is inversely proportional to the square of the distance). Therefore, the repulsive force in this scenario would be 15 N/2.56 = 5.86 N.

In the second problem, we are given two charged particles with a force of F between them, and we are asked to find the force when the distance between them is doubled. Using Coulomb's Law, we can calculate the new force by dividing the original force by 4 (as the distance is doubled, the force is inversely proportional to the square of the distance). Therefore, the force in this scenario would be F/4.
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of physics that describes the repulsive force between two electrically charged particles. It states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

2. What is the repulsive force?

The repulsive force is a force that acts between two particles with the same charge, causing them to push away from each other. It is the result of the particles' electric fields interacting with each other.

3. How is Coulomb's Law calculated?

Coulomb's Law is calculated using the following equation: F = k * (q1 * q2) / d^2, where F is the repulsive force, k is the proportionality constant, q1 and q2 are the charges of the particles, and d is the distance between them.

4. What is the relationship between distance and repulsive force?

The relationship between distance and repulsive force is inverse-square. This means that as the distance between two charged particles increases, the repulsive force between them decreases exponentially.

5. How does Coulomb's Law relate to the behavior of electrons in atoms?

Coulomb's Law plays a crucial role in explaining the behavior of electrons in atoms. The repulsive force between negatively charged electrons and the positively charged nucleus is responsible for keeping the electrons in orbit around the nucleus. This balance of forces is what allows atoms to exist as stable structures.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
752
  • Introductory Physics Homework Help
Replies
2
Views
898
  • Introductory Physics Homework Help
Replies
4
Views
656
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
503
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
2K
Back
Top