what is the value of the charge that experiences a force of 2.4×10⁻³ n in an electric field of 6.8×10⁻⁵…

Explanation:

Step1: Recall the formula

The formula relating electric force $F$, electric field $E$ and charge $q$ is $F = qE$. We need to solve for $q$, so $q=\frac{F}{E}$.

Step2: Substitute the given values

Given $F = 2.4\times10^{-3}\text{ N}$ and $E=6.8\times 10^{-5}\text{ N/C}$. Then $q=\frac{2.4\times 10^{-3}}{6.8\times 10^{-5}}$.

Step3: Calculate the value of $q$

Using the rule of exponents $\frac{a^m}{a^n}=a^{m - n}$, we have $q=\frac{2.4}{6.8}\times10^{-3-(-5)}=\frac{2.4}{6.8}\times10^{2}$. $\frac{2.4}{6.8}\approx0.35$, and $0.35\times 10^{2}=3.5\times 10^{1}\text{ C}$.

Answer:

D. $3.5\times 10^{1}\text{ C}$