Question

two charged particles are separated by a distance of 0.50 meters. the charge of one particle is 5.0×10⁻⁹ c and the charge of the other particle is - 1.0×10⁻⁷ c. which expression shows the value of the force between the two particles? coulombs law states that the electric force between two charged particles is given by the equation f = k\frac{q_1q_2}{r^2}, where f is the electric force, k is the coulomb constant (8.99×10⁹ n\frac{m^2}{c^2}), q_1 and q_2 are the charges of the particles (in coulombs), and r is the distance between the charged particles (in meters). a \frac{(8.99×10⁹)(5.0×10⁻⁹)( - 1.0×10⁻⁷)}{0.50^2} b \frac{(8.99×10⁹)(5.0×10⁻⁹)( - 1.0×10⁻⁷)}{(0.50)^2} c \frac{(8.99×10⁹)(5.0×10⁻⁹)( - 1.0×10⁻⁷)}{0.50^2} d \frac{(8.99×10⁹)(0.50)^2}{(5.0×10⁻⁹)( - 1.0×10⁻⁷)}

Explanation:

Step1: Identify the Coulomb's law formula

The electric - force formula is $F = k\frac{q_1q_2}{r^{2}}$, where $k = 8.99\times10^{9}\ Nm^{2}/C^{2}$, $q_1$ and $q_2$ are the charges of the two particles, and $r$ is the distance between them.

Step2: Substitute the given values

Given $q_1 = 5.0\times10^{-9}\ C$, $q_2=- 1.0\times10^{-7}\ C$, $r = 0.50\ m$, and $k = 8.99\times10^{9}\ Nm^{2}/C^{2}$. Substituting these values into the formula $F = k\frac{q_1q_2}{r^{2}}$, we get $F=\frac{(8.99\times10^{9})(5.0\times10^{-9})(-1.0\times10^{-7})}{0.50^{2}}$.

Answer:

C. $\frac{(8.99\times10^{9})(5.0\times10^{-9})(-1.0\times10^{-7})}{0.50^{2}}$