Question

Charges A and B are separated by a distance. If the distance is reduced to ½ the original value, what effect will it have on the electrostatic force?

Answer 1

B. Quadrupled.

Explanation:

By means of Coulomb's law we have the following:

`F=k\cdot(q_1\cdot q_2)/(r^2)`

Now, if the distance is halved (r' = 1/2 r), we substitute:

`\begin{gathered} F^(\prime)=k\cdot(q_1\cdot q_2)/(((1)/(2)r)^2) \\ \\ F^(\prime)=k(q_1q_2)/((1)/(4)r^2) \\ \\ F^(\prime)=4\cdot k\cdot(q_1\cdot q_2)/(r^2) \\ \\ F^(\prime)=4F \end{gathered}`

Which means that the force is quadrupled.

The correct answer is B. Quadrupled.