Question

In thefigure, the particles have charges q1-02 electrostatic force on particle 3? 230 nC and a3-4 98 nC, and distance a 5.1cm. What are the (a) x and (b) y components of the net

Answer 1

Electrostatic force is given by:
F = k*q1*q2/r^2

force is attractive if both charge have same signs, and force is repulsive if both charge have same sign.
Force on particle 3, due to charge at 4 will be in +x-direction
where, q1 = charge at 4 = -98 nC
q2 = +98 nC

& r = a = 5.1 cm = 5.1*10^-2 m

So,
F1 = (9*10^9)*[(98*10^-9)^2]/(5.1*10^-2)^2

F1 = 0.033232 N
Similarly, Force on particle 3, due to charge at 1 will be in -y-direction
where, q1 = charge at 1 = 230 nC
q2 = 98 nC

& r = a = 5.1*10^-2 m, So
F2 = (9*10^9)[(98*10^-9)(230*10^-9)]/(5.1*10^-2)^2

F2 = 0.077993 N
Now Force on particle 3 due to charge at 2 will be at 45 deg from +x-axis
where q1 = Charge at C = -230 nC
q2 = 98 nC and r = sqrt(2)*a = sqrt 2)*5.1*10^-2,

r = 7.21*10^-2 m

So
F3 = (9*10^9)[(98*10^-9)(230*10^-9)]/(7.21*10^-2)^2

F3 = 0.039023 N

Now Net force in x direction will be
Fx = F1 + F3*cos 45 deg
Fx = 0.033232  + 0.039023*cos 45 deg
Fx = 0.060825 N = 60.8*10^-3 N
Similarly, Now Net force in y direction will be
Fy = F2 + F3*sin 45 deg
Fy = -0.077993 + 0.039023*sin 45 deg
Fy = -0.050399 N = -50.4*10^-3 N
Now Net force will be
Fnet = Fx + Fy
Fnet = 0.060825 i - 0.050399 j

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