One particle has a charge of 2.15 x 10^ -9 while another particle has a charge of 3.22 x 10^ -9 If the two particles are separated by 0.015 m, what is the electromagnetic force between them? A. 4.31 x 10^ -7 N B. 2.77 x 10^ -4 N C. 4.15 x 10^ -6 N D. 6.22 x 10^ -4 N

Answer :

Answer:

B. 2.77 x [tex]10^{-4}[/tex] N

Explanation:

The required force can be calculated by:

F = [tex]\frac{Kq_{1}q_{2} }{d^{2} }[/tex]

Where F is the force between the particles, K is the coulomb's constant (9 x [tex]10^{9}[/tex] [tex]Nm^{2}/C^{2}[/tex]), [tex]q_{1}[/tex] is the charge on the first particle, [tex]q_{2}[/tex] is the charge on the second particle and [tex]d^{2}[/tex] is the distance between the particles.

So that:

F = [tex]\frac{9*10^{9}*2.15*10^{-9} *3.22*10^{-9} }{(0.015)^{2} }[/tex]

  = [tex]\frac{6.2307*10^{-8} }{(2.25*10^{-4} }[/tex]

  = 2.7692 x [tex]10^{-4}[/tex]

 The force between the particles is 2.77 x [tex]10^{-4}[/tex] N.

Answer:

b

Explanation:

a pex

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