Charges of 5µC, 10µC and -10µC are placed in air . . .

Question : Charges of 5µC, 10µC and -10µC are placed in air at the corners A, B and C of an equilateral triangle ABC of side 5 cm. Determine the resultant force on the charge at A.

Doubt by Lakshita

Solution : 

q_A=+5µC = +5×10^-6 C
q_B=+10µC = +10×10^-6 C
q_C=-10µC = -10×10^-6 C

a = 5 cm  = 5×10^-2 m

Charge +5µC at A will experience a repulsive force F_AB due to charge +10µC at point B

F_AB = kq_Aq_B/a² (Along BA)

Charge +5µC at A will experience a attractive force F_AC due to charge -10µC at point C

F_AC = kq_Aq_C/a² 
F_AC = kq_Aq_B/a² [∵|q_B| = |q_C| ]

F_AC = kq_Aq_B/a² (Along AC)

It is clear that 
F_AB=F_AC= kq_Aq_B/a² = F (Let)

Now, the angle (θ) between the direction of force FAB and F_AC is 120° so the magnitude of the resultant Force at A will be given by 

F_NET = √(F_AB²+F_AC²+2F_AB.F_ACcos120°)
F_NET = √[F²+F²+2F.F(-1/2)]
F_NET = √[2F²-F²]
F_NET = √F²
F_NET = F
F_NET = kqAqB/a²
F_NET = [(9×10⁹)×(5×10^-6)(10×10^-6)]/[5×10^-2]²
F_NET = [450×10^-3]/[25×10^-4]
F_NET = [4500×10^-4]/[25×10^-4]
F_NET = 4500/25
F_NET = 180 N

Hence, the magnitude of the net force on the charge at A will be 180 N

Now, the direction of the resultant force on charge at A will be given by 

tanβ = F_ACsinθ/(F_AB+F_ACcosθ)
tanβ = Fsin120°/(F+Fcos120°)
tanβ = F(√3/2)/[F+F(-1/2)]

[∵ sin120°=√3/2 & cos120°=-1/2]
tanβ = F(√3/2)/F/2
tanβ = √3
tanβ = tan60°
β = 60°

Hence, the resultant force vector is making an angle of 60° with the F_AC Vector.