Two charged particles exert an electrostatic force of 8 N on each other. What will the magnitude of the electrostatic force be if the distance between the two charges is REDUCED to one-third of the original distance?
Would you just have to multiply 8 by 9?
Nine. (3 squared)
To determine the magnitude of the electrostatic force when the distance between two charged particles is reduced, we need to understand how the electrostatic force is related to distance.
The electrostatic force between two charged particles is given by Coulomb's Law, which states that the force (F) is directly proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance between them (r):
F = k * (q1 * q2) / r^2
Where:
F is the magnitude of the electrostatic force,
k is the electrostatic constant,
q1 and q2 are the charges of the particles, and
r is the distance between the charges.
In this case, the electrostatic force between the charged particles is 8 N when the distance between them is at the original value. Let's call this distance r1.
Now, if we reduce the distance to one-third of r1, the new distance would be (1/3) * r1, or r2 = r1/3.
To find the new magnitude of the electrostatic force (F2), we substitute the new distance (r2) into Coulomb's Law:
F2 = k * (q1 * q2) / (r2)^2
F2 = k * (q1 * q2) / (r1/3)^2
Simplifying further:
F2 = k * (q1 * q2) * (3/r1)^2
F2 = k * (q1 * q2) * 9/r1^2
Since k, q1, and q2 remain constant, we can say that the new magnitude of the electrostatic force (F2) is directly proportional to 9/r1^2.
Therefore, to find the new magnitude, we multiply the original force (8 N) by 9/r1^2.
In conclusion, multiplying 8 N by 9 would give you the magnitude of the electrostatic force when the distance between the two charges is reduced to one-third of the original distance.