The force of repulsion that two like charges exert on each other is 3.0 N. What will the force be if the distance between the charges is increased to 3 times its original value?

You must have been told that the electrostatic force between two charges varies inversely with the square of the distance between them. That means, in this case, that the force gets multiplied by a factor 1/3^2 = 1/9.

To find out the force between two charges when the distance between them is increased, we can use Coulomb's Law.

Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be represented as:

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

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

In this case, we are given that the force of repulsion, F, is 3.0 N. Let's assume the initial distance between the charges is d.

Using Coulomb's Law, we can write:

3.0 N = k * (q1 * q2) / d^2

Now, we are asked to find the force when the distance is increased to 3 times its original value. Let's call this new distance 3d.

We need to determine the new force, F'.

Using Coulomb's Law again, we have:

F' = k * (q1 * q2) / (3d)^2
= k * (q1 * q2) / (9d^2)
= (1/9) * (k * (q1 * q2) / d^2)
= (1/9) * F

So, the force between the two charges when the distance is increased to 3 times its original value is 1/9 times the initial force. Therefore, the new force (F') will be:

F' = (1/9) * 3.0 N
= 0.33 N (approximately)

Therefore, the force will be approximately 0.33 N when the distance between the charges is increased to 3 times its original value.