Two point charges exert a 2.50 N force on each other. What will the force become if the distance between them is increased by a factor of 5?
1/50
To determine the new force between the two point charges when the distance between them is increased by a factor of 5, we can use Coulomb's Law.
Coulomb's Law states that the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Mathematically, Coulomb's Law can be expressed as:
F = k * (q1 * q2) / r^2
Where:
F = Force between the point charges
k = Coulomb's constant (approximately 9 × 10^9 N m^2/C^2)
q1, q2 = Magnitudes of the point charges
r = Distance between the point charges
We are given that the force between the point charges is 2.50 N. Let's assume this force is F1 and the initial distance between the charges is r1.
F1 = k * (q1 * q2) / r1^2
Now, we need to find the force F2 when the distance between the charges is increased by a factor of 5. Let's call the new distance r2.
We know that the relationship between r1 and r2 is given by:
r2 = 5 * r1
To find F2, we can substitute this relationship into Coulomb's Law:
F2 = k * (q1 * q2) / r2^2
= k * (q1 * q2) / (5 * r1)^2
= k * (q1 * q2) / 25 * r1^2
= (1/25) * k * (q1 * q2) / r1^2
= (1/25) * F1
So, the new force F2 when the distance between the point charges is increased by a factor of 5 will be one-fifth (1/25) of the initial force F1.
Substituting the given force F1 = 2.50 N into the equation, we get:
F2 = (1/25) * 2.50 N
= 0.10 N
Therefore, the force between the charges will become 0.10 N when the distance between them is increased by a factor of 5.
Use Couloumb's law:
F=q1q2/d^2
We get force is inversely proportional to the square of the distance. Solve the rest yourself.