Two point charges are on the $x$-axis. One charge, $q_{1}=10.0 \mathrm{nC}$, is located at the origin, and the other charge, $\mathrm{q}_{2}=18.0 \mathrm{nC}$, is located at $\mathrm{x}=9.00 \mathrm{~m}$. What is the force on $\mathrm{q}_{2}$ ?

Question

Quizplus · Accepted Answer

The force between two point charges is given by Coulomb's law: $F = k \frac{|q_1 q_2|}{r^2}$, where $k = 8.987 \times 10^9 \, \text{N m}^2/\text{C}^2$, $q_1 = 10.0 \, \text{nC} = 10.0 \times 10^{-9} \, \text{C}$, $q_2 = 18.0 \, \text{nC} = 18.0 \times 10^{-9} \, \text{C}$, and $r = 9.00 \, \text{m}$. Plugging in these values, we find $F = 20.0 \, \text{nN}$ in the direction from $q_1$ to $q_2$, which is the positive $x$-direction since $q_1$ is at the origin and $q_2$ is at a positive $x$ value.

Two Point Charges Are on The xxx -Axis One Charge q1=10.0nCq_{1}=10.0 \mathrm{nC}q1​=10.0nC

Two Point Charges Are on The $x$ -Axis One Charge $q_{1}=10.0 \mathrm{nC}$