Two equal charges repel one another with a force of 4.0 x 10-4 N when they are 10 cm apart. If they are moved until the separation is 20 cm, the repulsive force will be??
this is an inverse-square relation
the force is inversely proportional to the square of the distance
To determine the repulsive force when the charges are moved to a new separation distance, you can use Coulomb's law. Coulomb's law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The mathematical formula for Coulomb's law is:
F = (k * q1 * q2) / r^2
Where:
F is the force between the charges
k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2)
q1 and q2 are the magnitudes of the two charges
r is the separation distance between the charges
Now, let's calculate the repulsive force at the new separation distance:
Given:
Force at the initial separation (F1) = 4.0 x 10^-4 N
Initial separation distance (r1) = 10 cm = 0.1 m
New separation distance (r2) = 20 cm = 0.2 m
We can use the ratio of the inverse squares of the distances to determine the relationship between the two forces:
(F2/F1) = (r1^2/r2^2)
Using this relationship, we can calculate the force at the new separation distance:
F2 = F1 * (r1^2/r2^2)
F2 = (4.0 x 10^-4 N) * ((0.1 m)^2 / (0.2 m)^2)
F2 = (4.0 x 10^-4 N) * (0.01 / 0.04)
F2 = (4.0 x 10^-4 N) * 0.25
F2 = 1.0 x 10^-4 N
Therefore, the repulsive force when the separation is 20 cm is 1.0 x 10^-4 N.