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Two charges that are 2 meters apart repel each other with a force of 2 x 10-5 newton. If the distance between the charges is
decreased to 1 meter, the force of repulsion will be
A) 1 x 10-5 N B) 4 x 10-5 N C) 8 x 10-5 N D) 5 x 10-6 N

Respuesta :

nobillionaireNobley
The force of attraction / repultion of two charges is given by F = KQ1Q2 / d^2; where K is the Coulumbs constant, Q1 is the charge on charge 1, Q2 is the charge on charge 2, d is the distance of the two chardes apart.

Thus,
KQ1Q2 / 4 = 2 x 10^-5
KQ1Q2 = 4 x 2 x 10^-5 = 8 x 10^-5

When the distance between the charges was reduced to 1 meter.
F = KQ1Q2 / d^2 = 8 x 10^-5 / 1^2 = 8 x 10^-5

Therefore, option C is the correct answer.
samanthatorres240

Answer:

it is C.

Explanation:

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