Question

5 of 12
if the magnitude of each charge was doubled, then the force on each charge would be
a four times as big
b twice as big

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is \(F = k\frac{q_1q_2}{r^{2}}\), where \(k\) is the Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them.

Step2: Analyze the change when charges are doubled

Let the original charges be \(q_1\) and \(q_2\), and the new charges be \(Q_1 = 2q_1\) and \(Q_2=2q_2\). The new force \(F'=k\frac{Q_1Q_2}{r^{2}}=k\frac{(2q_1)(2q_2)}{r^{2}}\).
Simplify \(F'\): \(F' = 4k\frac{q_1q_2}{r^{2}}\). Since \(F = k\frac{q_1q_2}{r^{2}}\), we have \(F'=4F\).

Answer:

A. four times as big