Question

two charges that are 2 meters apart repel each other with a force of 2 × 10^{-5} newton. if the distance between the charges is decreased to 1 meter. the

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is \(F = k\frac{q_1q_2}{r^{2}}\). Let the initial force \(F_1=2\times 10^{-5}\space N\) and initial distance \(r_1 = 2\space m\), the final distance \(r_2=1\space m\). Since \(k\), \(q_1\) and \(q_2\) remain constant, we can use the ratio \(\frac{F_1}{F_2}=\frac{r_2^{2}}{r_1^{2}}\).

Step2: Calculate the final force

Substitute the values into the ratio formula. \(\frac{2\times 10^{-5}}{F_2}=\frac{1^{2}}{2^{2}}\). Cross - multiply gives \(F_2=(2\times 10^{-5})\times\frac{2^{2}}{1^{2}}\). \(F_2 = 8\times 10^{-5}\space N\).

Answer:

\(8\times 10^{-5}\space N\) (the first option)