Question

how does doubling the distance between the test charge and point charge affect the electrostatic force? a. the force doubles. b. the force quadruples. c. the force reduces by one - half. d. the force reduces by one - fourth.

Explanation:

Step1: Recall Coulomb's law

The electrostatic force $F$ between two charges is given by $F = k\frac{q_1q_2}{r^{2}}$, where $k$ is the electrostatic - constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Consider the new distance

Let the original distance be $r$ and the new distance $r'=2r$. The new force $F'$ is $F' = k\frac{q_1q_2}{(2r)^{2}}$.

Step3: Simplify the new - force expression

$F' = k\frac{q_1q_2}{4r^{2}}=\frac{1}{4}\times k\frac{q_1q_2}{r^{2}}$. Since $F = k\frac{q_1q_2}{r^{2}}$, we have $F'=\frac{1}{4}F$.

Answer:

D. The force reduces by one - fourth.