Question

1. if you double the distance between two objects, the electric force is ______. (1 point) one half as strong quadrupled one fourth as strong doubled

Explanation:

Step1: Recall Coulomb's Law

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

Step2: Analyze the effect of doubling the distance

Let the initial distance be \(r_1 = r\), and the initial force be \(F_1=k\frac{q_1q_2}{r^{2}}\). When the distance is doubled, \(r_2 = 2r\). Then the new force \(F_2=k\frac{q_1q_2}{(2r)^{2}}=k\frac{q_1q_2}{4r^{2}}\).

Step3: Compare \(F_1\) and \(F_2\)

We can see that \(F_2=\frac{1}{4}F_1\).

Answer:

one fourth as strong