select the correct answer.\nthe force of attraction between two like - charged table tennis balls is…

Explanation:

Step1: Write Coulomb's law formula

$F = k\frac{q_1q_2}{r^2}$, where $F$ is the force, $k$ is the Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance. We need to solve for $r$.

Step2: Rearrange the formula for $r$

$r^2=k\frac{q_1q_2}{F}$, then $r = \sqrt{k\frac{q_1q_2}{F}}$.

Step3: Substitute the given values

$q_1 = 3.8\times 10^{-8}\text{ C}$, $q_2 = 3.0\times 10^{-8}\text{ C}$, $F=2.4\times 10^{-5}\text{ N}$, $k = 9.0\times 10^{9}\text{ N}\cdot\text{m}^2/\text{C}^2$. $r=\sqrt{\frac{9.0\times 10^{9}\times(3.8\times 10^{-8})\times(3.0\times 10^{-8})}{2.4\times 10^{-5}}}$ First, calculate the numerator: $9.0\times 10^{9}\times(3.8\times 10^{-8})\times(3.0\times 10^{-8})=9.0\times3.8\times3.0\times10^{9 - 8-8}=102.6\times 10^{-7}=1.026\times 10^{-5}$. Then, $r=\sqrt{\frac{1.026\times 10^{-5}}{2.4\times 10^{-5}}}=\sqrt{0.4275}\approx 0.65\text{ m}$.

Answer:

D. 0.65 meters