If two charges are a distance R apart from eachother and their force is F, what will the force be if they are moved apart to a distance of 2R?

F=Ko(Q1*Q2/r^2)

if r is doubled, the F will be reduced by 1/4.

opps, it will be 1/4 of the original F

To calculate the new force between two charges when they are moved apart, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Coulomb's Law equation:

F = k * (q1 * q2) / r^2

Where:
F is the force between the charges,
k is the Coulomb's constant (approximately 9 x 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
r is the distance between the charges.

In this case, if the charges are initially separated by a distance R and the force between them is F, we can write the equation as:

F = k * (q1 * q2) / R^2

Now, we are asked to find the new force when the charges are moved apart to a distance of 2R. Let's calculate it.

For the new distance of 2R, we can substitute r with 2R in the Coulomb's Law equation:

F' = k * (q1 * q2) / (2R)^2
F' = k * (q1 * q2) / 4R^2

Simplifying, we have:

F' = F / 4

Therefore, the force will be one-fourth (1/4) of the original force when the charges are moved apart to a distance of 2R.