Question

Two small spheres, A and B, are separated by a distance of 1.5 m as shown below. Sphere A has a charge of 3 μC and Sphere B has a charge of -6 μC. (The prefix \"μ\" means \"micro\" which is a \"millionth.\" So, 1 microCoulomb = one-millionth of a Coulomb = 10-6 C.)

Sphere C, is now moved to a new position that is directly above Sphere A. The distance between A and C is still half the distance between A and B

What is the magnitude of the NET electrical force exert on Sphere A due to B and C?

C, having a charge -3 μC

F_A,net =

Answer 1

See the diagram :

F1 = force exerted by A on C =

F2 = force exerted by B on C =

where k = 9 x 109 Nm2C-2.

Note : I have given the direction of the forces, so I will not use sign of charge in the calculations.

given :

q1 = 3 = 3 x 10-6 C

q2 = -6 = - 6 x 10-6 C

q3 = -3 = -3 x 10-6 C

r = 1.5 m.

Applying parallelogram law of addition to calculate net force :

=> Fnet = 128.98 x 10-3 N. [Answer]