Answer:
a.[tex]F=1.03\times 10^{-7}N[/tex]
b.Repulsive
Explanation:
We are given that
[tex]q_1=7.45nC=7.45\times 10^{-9}C[/tex]
[tex]1nC=10^{-9}C[/tex]
[tex]q_2=4.22nC=4.22\times 10^{-9}C[/tex]
r=1.66m
We know that
Electrostatic force =[tex]F=k\frac{q_1q_2}{r^2}[/tex]
r=Distance between [tex]q_1\;and\;q_2[/tex]
k=Constant=[tex]9\times 10^9Nm^2C^{-2}[/tex]
Using the formula
a.The magnitude of the electrostatic force=[tex]F=\frac{9\times 10^9\times 7.45\times 10^{-9}\times 4.22\times 10^{-9}}{(1.66)^2}[/tex]
The magnitude of the electrostatic force=[tex]F=1.03\times 10^{-7}N[/tex]
b.Both charge are positive .We know that when like charges repel each other and unlike charges attract to each other.
Therefore, the force between given charges is repulsive.
