In the figure what are the a magnitude and b direction from

In the figure what are the (a) magnitude and (b) direction (from x-axis s In the counterclockwise direction) of the net electrostatic force on particle 4 due to the other three particles? All four particles are fixed in the xy plane, and q_1 = -22.40 x 10^19 C, q_2 = +22.40 x 10^-19 C, Q_3=+44.60 x 10^-19 C, q_4 = +22.40 x 10^-39C, = 48degree, d_1 = 4.40 cm, and d_2 - d_3 = 2.80 cm.

Solution

The magnitude of force F₁₄ is

F₁₄=K|Q₁||Q₄|/d₁² = (9*10⁹)(22.40*10^-19)²/0.044²

F₁₄=2.33*10^-23 N ,and the angle O₁₄=180+48 =228^o

The magnitude of force F₂₄ is

F₂₄=K|Q₂||Q₄|/d₁² = (9*10⁹)(22.40*10^-19)²/0.028²

F₂₄=5.76*10^-23 N ,and the angle is O₂₄ =270^o

The magnitude of force F₃₄ is

F₃₄ =K|Q₃||Q₄|/d₃² =(9*10⁹)(22.4*10^-19)(44.8*10^-19)/0.028²

F₃₄ =1.152*10^-22 N ,at an angle O₃₄=180^o

Net force on in x-direction

F_net,x=F_1x+F_2x+F_3x = F₁₄Cos228-F₃₄ =-1.3*10^-22 N

Net force on in y-direction

F_net,y=F_1y+F_2y+F_3y = F₁₄sin228-F₂₄ =-7.5*10^-23 N

a)

Magnitude of Net force

F_Net=sqrt[F_x²+F_y²]=sqrt[(-1.3*10^-22)²+(-7.5*10^-23)²]

F_Net=1.5*10^-22 N

b)

Direction

O=tan^-1(F_y/F_x) =tan^-1(-7.5*10^-23/-1.3*10^-22)

O=30^o

From x-axis in counterclockwise direction is

O=30+180 =210^o