Two charges attract each other with a force of 3.0 N. What will be the force if the distance between them is reduced to one-ninth of its original value?
would it go up by 1/(1/9)^2 or 81 times?
what is 3x81N
To determine the new force between the charges when the distance is reduced, we can use Coulomb's Law equation:
F = (k * q1 * q2) / d^2
where:
F is the electrostatic force between the charges,
k is the electrostatic constant (9.0 x 10^9 N*m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
d is the distance between the charges.
Let's assume the original distance between the charges is d, and the original force between them is F1.
Given:
F1 = 3.0 N
d = original distance
If we reduce the distance to one-ninth of its original value, the new distance between the charges (d') will be:
d' = (1/9) * d = d/9
Now we can calculate the new force (F') using the new distance:
F' = (k * q1 * q2) / (d/9)^2
F' = (k * q1 * q2) / (d^2 / 81)
F' = (k * q1 * q2) * (81 / d^2)
F' = 81 * (k * q1 * q2) / d^2
Since the charges remain the same, we can express the new force in terms of the original force as follows:
F' = 81 * F1
Therefore, when the distance between the charges is reduced to one-ninth of its original value, the new force will be 81 times the original force.