Particles of charge + 65, + 48, and?95? C are placed in a line (Figure 1). The center one is L = 40cm from each of the others.

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a) Calculate the net force on the left charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left. Express your answer to two significant figures and include the appropriate units.

b) Calculate the net force on the center charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left. Express your answer to two significant figures and include the appropriate units.

c) Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left. Express your answer to two significant figures and include the appropriate units.

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the + 65μC charge, we consider it to be the origin. Hence, the two other charges are on the + x axis.

The net coulombs force on the charge is

F = [KQ (1) Q (2) ] / (r^2) + [KQ (1) Q (3) ] / (r^2)

Where K = Coloumbs constant =

Q (1) = charge on the leftmost side.

Q (2) = charge in the middle.

Q (3) = charge on the rightmost side.

F = [ (8.988 * 10^9) * (65*10^-6) * (48*10^-6) ] / (40^2) + [ (8.988 * 10^9) * (-95*10^-6) * (65*10^-6) ] / (40^2)

F = 0.01753 - 0.03469

F = - 0.017N

It has a negative sign, hence, it acts to the left.

B. For the + 48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the + x axis.

The net coulombs force on the charge is

F = [-KQ (1) Q (3) ] / (r^2) + [KQ (2) Q (3) ] / (r^2)

F = [ - (8.988*10^9) * (65*10^-6) * (48*10^-6) ] / (40^2) + [ (8.988 * 10^9) * (48*10^-6) * (-95*10^-6) ] / (40^2)

F = - 0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the - 95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ (1) Q (3) ] / (r^2) - [KQ (2) Q (3) ] / (r^2)

F = [ - (8.988*10^9) * (65*10^-6) * (-95*10^-6) ] / (40^2) - [ (8.988 * 10^9) * (48*10^-6) * (-95*10^-6) ] / (40^2)

F = + 0.03469 + 0.02562

F = + 0.060N

It has a positive sign, hence, it acts to the right.