If the two equal charges separated with one unit (one meter) has a force of 20 N. What will be the force if the distance is increased by 2 units (two meters)? What is the value of the new force compared to the old?
1 + 2 = 3 m is the final separation
R^2 increases by a factor of 9.
The electrostatic repulsion force decreases by a factor of 9.
To find the force if the distance is increased by 2 units, we can make use of 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 expressed as:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges
k is the electrostatic constant (approximate value is 8.99 x 10^9 Nm^2/C^2)
q1 and q2 are the charges
r is the distance between the charges
In this case, the charges are equal, so let's assume they are both q.
Given:
F1 = 20 N (force when the distance is 1 unit or 1 meter)
r1 = 1 unit (distance between the charges is 1 meter)
We need to find:
F2 (force when the distance is increased by 2 units)
r2 (new distance between the charges)
We know that the force is inversely proportional to the square of the distance. So, if the distance is increased by 2 units, the new distance would be r1 + 2.
Therefore, r2 = r1 + 2 = 3 units (distance between the charges is 3 meters).
Now we can calculate the new force (F2) using Coulomb's Law:
F2 = k * (q * q) / r2^2
Substituting the given values:
F2 = (8.99 x 10^9 Nm^2/C^2) * (q * q) / (3^2)
Simplifying,
F2 = (8.99 x 10^9 Nm^2/C^2) * (q^2) / 9
From the given information, we can conclude that F2 will be 20/9 N (approximately 2.22 N) and the new force compared to the old force is F2/F1 = (20/9 N) / (20 N) = 1/9 (approximately 0.111 or 11.1%).
So, the new force is about 11.1% of the old force.