If the distance between two charges is increased to three times the original distance, how will the electrical force between the charges compare with the original force?
It will increase to three times the original force.
It will increase to nine times the original force.
It will decrease to one-third the original force.
It will decrease to one-ninth the original force.
Decrease to one-ninth
Thank you :)
The electrical force between two charges is inversely proportional to the square of the distance between them.
If the distance between two charges is increased to three times the original distance, the force will decrease to one-ninth (1/9) of the original force.
To determine how the electrical force between charges changes with distance, we can use the inverse square law of electrostatics. According to this law, the electrical force is inversely proportional to the square of the distance between the charges.
Let's denote the original distance between the charges as "d" and the original force as "F." If the distance is increased to three times the original distance, the new distance would be "3d."
Using the inverse square law, we can calculate the ratio of the new force (F') to the original force (F) as follows:
(F')/(F) = (d^2)/[(3d)^2]
Simplifying the equation, we get:
(F')/(F) = (1/d^2)/(1/[3^2 * d^2])
= (1/d^2)/(1/9d^2)
= 1/1 * 9/1
= 9
Therefore, the electrical force between the charges will increase to nine times the original force when the distance is increased to three times the original distance.
So, the correct answer is: "It will increase to nine times the original force."