Question

the amount of force between two positive charges is 40 newtons. if the charge stays the same but the distance is doubled, how many newtons will there be between the two charges?

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is \(F = k\frac{q_1q_2}{r^{2}}\), where \(F\) is the force between two charges, \(k\) is a constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them. Initially, \(F_1=k\frac{q_1q_2}{r_1^{2}} = 40\space N\).

Step2: Analyze the new - distance case

When the distance is doubled (\(r_2 = 2r_1\)), the new force \(F_2=k\frac{q_1q_2}{r_2^{2}}\). Substitute \(r_2 = 2r_1\) into the formula: \(F_2=k\frac{q_1q_2}{(2r_1)^{2}}=k\frac{q_1q_2}{4r_1^{2}}\).

Step3: Relate \(F_2\) to \(F_1\)

Since \(F_1 = k\frac{q_1q_2}{r_1^{2}}\), then \(F_2=\frac{F_1}{4}\). Given \(F_1 = 40\space N\), so \(F_2=\frac{40}{4}\).

Answer:

\(10\space N\)