Question

A charge of -2.95 µC is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass, which has a radius of 0.105 m. The charges on the circle are -4.4 µC at the position due north and +5.35 µC at the position due east. What are the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east. magnitude in N, direction in Degrees

Answer 1

Based on Coulomb's Law the force between two charged particles is:

The Force of charge due north is: (9*10⁹Nm²/C²)(-2.95*10^-6C)(-4.4*10^-6C)/(.105)² = 10.6 N going south because of like charges.( repulsion)

The Force of charge due east is: (9*10⁹Nm²/C²)(-2.95*10^-6C)(5.35 *10^-6C)/(.105)² = 12.88 N going east beacsue of unlike charges ( attraction).

So the magnitude of the force = √[(10.6)² + (12.88)²] = 16.68 N

now find the angle: tan^-1(10.6/12.88) = 39.45^o

Hence , the direction is south of east.