Question

Two charges and 2.0 C and 13 C are placed 4.0 m apart. What is the force between these charges?

Answer 1

`1.460875\cdot10^10 N`

Explanation:

Coulomb's Law is can be dictated by:
`F= k \frac(q_1q_2)(r^2)`
where the Coulomb's law constant `k=\frac(1)(4\pi\epsilon_0)=8.99\cdot10^9 \frac(Nm^2)(C^2)`

Therefore, the force between these charges is:
`F= (8.99\cdot10^9) \frac((2\cdot13))(4^2)=1.460875\cdot10^10 N`

Also, since both charges are positively charged, you expect a repulsive force between the charges.