Respuesta :
The electrostatic force between the charges is 11.25 x [tex]10^{-11} \;\mu N[/tex].
We have two charges separated by some distance apart.
We have to calculate the magnitude of the force exerted on the charges due to the electromagnetic interaction in μN (10-6 Newtons).
What is the formula to calculate the electrostatic force between two point charges ?
The force between two point charges is -
F = [tex]$\frac{1}{4\pi \epsilon_{o} } \frac{qQ}{r^{2} }[/tex]
According to the question -
Q =5 nC = 5 x [tex]10^{-9}[/tex] C
q = 1 nC = [tex]10^{-9}[/tex] C
r = 2 cm = 0.02 m
The force of repulsion between the charges is -
[tex]$\frac{1}{4\pi \epsilon_{o} } \frac{qQ}{r^{2} }[/tex]
[tex]$F =9\times 10^{9} \times \frac{5\times 10^{-9}\times 10^{-9} }{(0.02)^{2} }[/tex]
F = 11.25 x [tex]10^{-5}[/tex] Newtons.
Now , in micro - newtons →
F = 11.25 x [tex]10^{-5} \times 10^{-6}[/tex] = 11.25 x [tex]10^{-11} \;\mu N[/tex].
Hence, the electrostatic force between the charges is 11.25 x [tex]10^{-11} \;\mu N[/tex].
To solve more questions on Electrostatics, visit the link below-
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[ The given question is not complete. The complete question is -
" Two point-like particles separated by 2 cm have charges q1=5 nC and q2=1 nC. Note that the force felt by each charge due to the other is a vector and has both a magnitude and a direction. What is the magnitude of the force exerted on the charges due to the electromagnetic interaction, in μN (10-6 Newtons)? " ]
