What happens to the force between the charges if the distance between the charges is increased to two times its original value?
Value of the force is increased by two.
Value of the force is decreased by two.
Value of the force is decreased by four.
Value of the force is increased by four.
I would guess the obvious, which would be "value of the force is decreased by two."
Value of the force is decreased by four man!
Value of the force is decreased by four.
F=k•q1•q2/r²
If r2=2•r1, F2=F1/4
To determine the effect of increasing the distance between charges on the force between them, we can refer to Coulomb's Law.
Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Mathematically, the formula can be expressed as: F = k * (q1 * q2) / r^2
Where F represents the force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Now, if the distance between the charges is increased to two times its original value, we can assume that the new distance (r') is twice the original distance (r), i.e., r' = 2r.
Substituting this into Coulomb's Law, we get:
F' = k * (q1 * q2) / (2r)^2
= k * (q1 * q2) / 4r^2
= (1/4) * (k * (q1 * q2) / r^2)
As we can see from the equation, increasing the distance between the charges by a factor of two results in the force decreasing by a factor of four (1/4). Therefore, the correct answer is "Value of the force is decreased by four."