Particle 1 carrying -4.0 μC of charge is fixed at the origin of an xy coordinate system, particle 2 carrying +8.0 μC of charge is located on the x axis at x = 2.0 m , and particle 3, identical to particle 2, is located on the x axis at x = -2.0 m .

Question

Particle 1 carrying -4.0 μC of charge is fixed at the origin of an xy coordinate system, particle 2 carrying +8.0 μC of charge is located on the x axis at x = 2.0 m , and particle 3, identical to particle 2, is located on the x axis at x = -2.0 m .
Part A
What is the vector sum of the electric forces exerted on particle 3? Determine the x and y components of the vector sum.

Damon · Answer

All of your particles lie on the x axis. The force is therefore also along the x axis !
F = k Qa Qb /d^2, repulsive if same sign of Q
so
Fx at 3 = k (8 μC) (4 μC)/4 pulling right - k (8 μC) (8 μC)/16 pushing left
= k (+8-4) = +4 k μ^2

dineo · Answer

q=2.0*10^-6 C