Answer:
0.71 cm
Explanation:
[tex]q_1=75.0nC=75\times 10^{-9}C[/tex]
[tex]1nC=10^{-9}C[/tex]
[tex]q_2=75.0nC=75\times 10^{-9}C[/tex]
Force between two charges=1 N
Coulomb's law of force
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where k=[tex]9\times 10^9Nm^2/C^2[/tex]
Using the formula
[tex]1=\frac{9\times 10^9\times 75\times 10^{-9}\times 75\times 10^{-9}}{r^2}[/tex]
[tex]r^2=0.50625\times 10^{-4}[/tex]m
[tex]r=\sqrt{0.50625\times 10^{-4}}=0.71\times 10^{-2} m[/tex]
[tex]r=0.71\times 10^{-2}\times 10^{2}=0.71\times 10^{-2+2}=0.71\times 10^0=0.71 cm[/tex]
Using formula
[tex]1 m=10^2cm, a^x\cdot a^y=a^{x+y}, a^0=1[/tex]
Hence, the distance between two charges =r=0.71 cm
