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

1 Answer

#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.