Question

two charges that are both - 3.0 c push each other apart from 2 meters away. what is the force?

Explanation:

Step1: Identify the formula

Coulomb's law: \( F = k\frac{q_1q_2}{r^2} \), where \( k = 9\times10^{9}\ N\cdot m^{2}/C^{2}\), \(q_1 = q_2=- 3.0\ C\), \(r = 2\ m\)

Step2: Substitute the values into the formula

\(F=(9\times 10^{9})\frac{(-3.0)\times(-3.0)}{2^{2}}\)
First, calculate \((-3.0)\times(-3.0)=9.0\) and \(2^{2} = 4\)
Then \(F=(9\times 10^{9})\frac{9.0}{4}\)
\(F = 9\times10^{9}\times2.25\)
\(F=2.025\times 10^{10}\ N\)

Answer:

\(2.025\times 10^{10}\ N\)