Question

How strong is the repulsive force exerted on two point charges that each carry 1.0 E-6 C of negative charge and are 0.30 meters apart?Coulomb's Constant is 9.0 E9 N*m^2/C^2

Answer 1

0.1 N

Step-by-step explanation:

The magnitude of a force exerted by two charges q1 and q2 that are separated by a distance of r is calculated as

`F=k(q_1q_2)/(r^2)`

So, replacing k = 9 x 10^9 N m²/C², r = 0.30 m, and q1 and q2 by 1.0 x 10^-6 C, we get

`\begin{gathered} F=(9*10^9Nm^2/C^2)((1*10^(-6)C)(1*10^(-6)C))/((0.30m)^2) \\ F=0.1N \end{gathered}`

Therefore, the repulsive force has a magnitude of 0.1 N