- You firstly must draw a diagram to solve problem.
![enter image source here]()
The magnitude of force is the same as the values are the same.
- Electric charges are on the corners of the isosceles triangle.
- You must perform the necessary unit conversions.
"50 cm=0.5 m"50 cm=0.5 m
+3 mu C=+3*10^-6C+3μC=+3⋅10−6C
+5 mu C=+5*10^-6C+5μC=+5⋅10−6C
-5 mu C=-5*10^-6C−5μC=−5⋅10−6C
sin alpha=40/50=4/5sinα=4050=45
cos alpha=30/50=3/5cosα=3050=35
- We know that electrical charges force each other.
F=k*(q_1*q_2)/r^2F=k⋅q1⋅q2r2
- Pushing force occurs between the electric charge at point A and the electric charge at point B.
![enter image source here]()
color(blue)(F_(BA)=k*(q_A*q_B)/(0.5)^2=(9*10^9*5*10^-6*3*10^-6)/(0.25)=(135*10^-3)/0.25=540*10^-3=0.54N)FBA=k⋅qA⋅qB(0.5)2=9⋅109⋅5⋅10−6⋅3⋅10−60.25=135⋅10−30.25=540⋅10−3=0.54N
- A pulling force occurs between the electric charge at point A and the electric charge at point C.
![enter image source here]()
- The magnitude of force is the same as the values are the same.
color(green)(F_(CA)=-0.54N)FCA=−0.54N
- We must find the vector sum of the blue and green forces.
![enter image source here]()
F=sqrt(F_(BA)^2+F_(CA)^2+2*F_(BA)*F_(CA)*cos(2 alpha)F=√F2BA+F2CA+2⋅FBA⋅FCA⋅cos(2α)
cos(2alpha)=cos^2 alpha-sin^2 alphacos(2α)=cos2α−sin2α
cos(2alpha)=(3/5)^2-(4/5)^2=9/25-16/25=-7/25cos(2α)=(35)2−(45)2=925−1625=−725
F=sqrt((0.54)^2+(0.54)^2+2*0.54*(0.54)*(-7/25))F=√(0.54)2+(0.54)2+2⋅0.54⋅(0.54)⋅(−725)
F=sqrt(0.2916+0.2916-2*0.2916*7/25)NF=√0.2916+0.2916−2⋅0.2916⋅725N
F=sqrt(0.5832-0.163296)F=√0.5832−0.163296
F=0.65 NF=0.65N
