select the correct answer.\nthe electric force between two charged objects is 0.24 newtons. if the distance…

Explanation:

Step1: Recall Coulomb's law

The electric - force formula 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. Let the initial force be $F_1=k\frac{q_1q_2}{r_1^{2}} = 0.24$ N.

Step2: Analyze the new - distance situation

The new distance $r_2 = 4r_1$. The new force $F_2=k\frac{q_1q_2}{r_2^{2}}=k\frac{q_1q_2}{(4r_1)^{2}}$.

Step3: Express $F_2$ in terms of $F_1$

$F_2=k\frac{q_1q_2}{16r_1^{2}}=\frac{1}{16}\times k\frac{q_1q_2}{r_1^{2}}$. Since $F_1 = k\frac{q_1q_2}{r_1^{2}} = 0.24$ N, then $F_2=\frac{F_1}{16}$.

Step4: Calculate the new force

$F_2=\frac{0.24}{16}=0.015$ N.

Answer:

A. 0.015 newtons