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

ANSWER:

B. Quadrupled.

STEP-BY-STEP EXPLANATION:

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

$F=k\cdot\frac{q_1\cdot q_2}{r^2}$

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

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

Which means that the force is quadrupled.

The correct answer is B. Quadrupled.