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3 years ago

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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 separat

ed 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

1 answer:

marin [14]3 years ago

3 0

Answer:

B. 2.77 x $10^{-4}$ N

Explanation:

The required force can be calculated by:

F = $\frac{Kq_{1}q_{2} }{d^{2} }$

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

So that:

F = $\frac{9*10^{9}*2.15*10^{-9} *3.22*10^{-9} }{(0.015)^{2} }$

= $\frac{6.2307*10^{-8} }{(2.25*10^{-4} }$

= 2.7692 x $10^{-4}$

The force between the particles is 2.77 x $10^{-4}$ N.

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Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

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