two egatively charged particles repel one other with an electrostatic force of 5 newtons. determine how the sign and manitude of this force changes when; a. the distance between them is doubled, b. the distance is cut in half., c. one particle's charge is doubled., d both particles charges are tripled.
I will be happy to critique your thinking.
Use Coulomb’s Law
F= k•q1•q2/r^2
The for the first question:
F1 = k•q1•q2/(2r)^2
Find the ratio
F1/F = (k•q1•q2)• r^2/(k•q1•q2)• (2r)^2 =1/4.
F1 =F/4=5/4=1.25 N .
….
Act like that for other cases.
To determine how the sign and magnitude of the electrostatic force changes under different conditions, we need to understand the relationship between the force, distance, and charge.
The electrostatic force between two charged particles is given by Coulomb's law:
F = k * (|q1 * q2|) / r^2
Where:
- F is the electrostatic force between the two particles
- k is the electrostatic constant (approximately 9 × 10^9 N m^2/C^2)
- q1 and q2 are the charges of the particles
- |q1 * q2| represents the product of the magnitudes of the charges
- r is the distance between the particles
Now let's consider each scenario and determine how it affects the force:
a. Doubling the distance:
If the distance between the particles is doubled (2r), then the force will decrease because of the inverse square relationship.
Let's calculate the new force using the formula:
F' = k * (|q1 * q2|) / (2r)^2
F' = k * (|q1 * q2|) / 4r^2
The new force is one-fourth (1/4) of the original force.
b. Halving the distance:
If the distance between the particles is halved (r/2), then the force will increase because of the inverse square relationship.
Let's calculate the new force:
F'' = k * (|q1 * q2|) / (r/2)^2
F'' = k * (|q1 * q2|) / (1/4r^2)
F'' = 4 * (k * |q1 * q2|) / r^2
The new force is four times (4x) the original force.
c. Doubling one particle's charge:
If the charge of one particle is doubled (2q1 or 2q2), then the force will double as well because the charge is directly proportional to the force.
F''' = k * (|2q1 * q2|) / r^2
F''' = 2 * (k * |q1 * q2|) / r^2
The new force is twice (2x) the original force.
d. Tripling both particles' charge:
If both particles' charges are tripled (3q1 and 3q2), then the force will increase by a factor of nine (3^2 = 9) because the charges are directly proportional to the force.
F'''' = k * (|3q1 * 3q2|) / r^2
F'''' = 9 * (k * |q1 * q2|) / r^2
The new force is nine times (9x) the original force.
So, to summarize, when:
a. The distance is doubled, the force becomes one-fourth (1/4) of the original force.
b. The distance is halved, the force becomes four times (4x) the original force.
c. One particle's charge is doubled, the force becomes twice (2x) the original force.
d. Both particle's charges are tripled, the force becomes nine times (9x) the original force.