Two charges q1 and q2 when separated by a distance r1 = R produce a force F1on each other. If the separation is increased to r2 = 2R, what is the new force F2 in terms of F1?
The Coulomb force between charges is inversely proportional to the square of distance between them.
If the distance doubles, the force gets multiplied by (1/2)^2 = 1/4
To find the new force F2 in terms of F1 when the separation is increased to r2 = 2R, we need to understand how the force between two charges varies with distance.
The force between two charges, known as the Coulomb's law, is given by:
F = k * (|q1| * |q2|) / r^2
where F is the force, q1 and q2 are the magnitudes of the charges, r is the separation between them, and k is the Coulomb's constant.
In this case, we have two charges q1 and q2 separated by distance r1 = R producing a force F1 on each other. We want to find the new force F2 when the separation is increased to r2 = 2R.
Let's assume the charges q1 and q2 are equal in magnitude but opposite in sign. Hence, q1 = q2 = q.
First, we need to determine how the force F1 depends on the separation r1. Since the force depends inversely on the square of the separation distance, we can write:
F1 = k * (|q| * |q|) / (r1^2)
= k * q^2 / R^2 (as r1 = R)
Now, we want to find the new force F2 in terms of F1 when the separation is increased to r2 = 2R.
Using Coulomb's law, we can write:
F2 = k * (|q| * |q|) / (r2^2)
= k * q^2 / (2R)^2
= k * q^2 / 4R^2
= (1/4) * (k * q^2 / R^2)
= (1/4) * F1
Therefore, the new force F2 is equal to one-fourth (1/4) of the original force F1.
In conclusion, when the separation between two charges is increased from r1 = R to r2 = 2R, the new force F2 is one-fourth (1/4) of the original force F1.