Question

question 13 of 25 what is the electric force between two point - charges when $q_1 = + 2e$, $q_2 = + 3e$, and $r = 0.05$ m? ($f_e=\frac{kq_1q_2}{r^2}$, $e = 1.6\times10^{-19}c$, $k = 9.00\times10^{9}ncdot m^{2}/c^{2}$) a. $1.1\times10^{-24}n$ b. $-1.1\times10^{-24}n$ c. $-5.5\times10^{-25}n$ d. $5.5\times10^{-25}n$

Explanation:

Step1: Substitute values into formula

Given $q_1 = + 2e$, $q_2=+3e$, $e = 1.6\times10^{-19}\text{ C}$, $k = 9.00\times10^{9}\text{ N}\cdot\text{m}^2/\text{C}^2$, $r = 0.05\text{ m}$. First, find $q_1$ and $q_2$ in Coulombs. $q_1=2\times1.6\times 10^{-19}\text{ C}=3.2\times10^{-19}\text{ C}$, $q_2 = 3\times1.6\times10^{-19}\text{ C}=4.8\times10^{-19}\text{ C}$. Then use Coulomb's law $F_e=\frac{kq_1q_2}{r^{2}}$.

Step2: Calculate the force

Substitute the values into the formula: $F_e=\frac{9.00\times 10^{9}\text{ N}\cdot\text{m}^2/\text{C}^2\times3.2\times10^{-19}\text{ C}\times4.8\times10^{-19}\text{ C}}{(0.05\text{ m})^{2}}$. First, calculate the numerator: $9.00\times10^{9}\times3.2\times10^{-19}\times4.8\times10^{-19}=9.00\times3.2\times4.8\times10^{9 - 19-19}=138.24\times10^{-29}\text{ N}\cdot\text{m}^2$. Then, calculate the denominator: $(0.05\text{ m})^{2}=0.0025\text{ m}^2$. Now, $F_e=\frac{138.24\times10^{-29}\text{ N}\cdot\text{m}^2}{0.0025\text{ m}^2}=55.296\times10^{-25}\text{ N}\approx5.5\times 10^{-25}\text{ N}$.

Answer:

D. $5.5\times 10^{-25}\text{ N}$