Three objects with charges +5.0 μC, -6.0 μC, and +7.0 μC are placed in a line, Determine the magnitude and direction of the net electric force on each charge.

Question

+5.0μC -6.0 μC +7.0 μC <---- charges

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0.80m 0.40m <---- distance between objects

Answer 1

The+ 5 μC charge will experience a force to the right due to the attractive force from the -6 μC charge and a force to the left due to the +7 μC charge. The net force on the first charge is the sum of the two forces, one positive and one negative.

The force to the right is F = k x (+5 x 10^-6 C )x (-6 x 10^-6 C) / 0.8² m². The force to the left is

F = k x (+5 x 10^-6 ) x (+7 x 10^-6C) / 1.2² m².

F_right = 8.99 x 10⁹ x (+5 x 10^-6 C )x (-6 x 10^-6 C) / 0.8² m² = - 0.42 N

F_left = 8.99 x 10⁹ x (+5 x 10^-6 ) x (+7 x 10^-6C) / 1.2² m² = 0.22 N

F_right + F_left = F_net = -0.42 N+ 0.22 N = - 0.2 N

The negative sign indicates that the net force on the +5.0 μC is attractive and to the right.

The same reasoning applies to the other two charges which you should be able to figure out.