Use k = 8,99 ⋅ 10 9 N \cdotp m 2 / C 2 k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2 k = 8 , 99 ⋅ 1 0 9 N \cdotp m 2 / C 2 . Two Coulomb forces act on charge q 3 q_3 q 3 : F ⃗ 13 \vv{F}_{13} F 13 (from q 1 q_1 q 1 ) and F ⃗ 23 \vv{F}_{23} F 23 (from q 2 q_2 q 2 ). Both have magnitude given by Coulomb’s law with r = ℓ r = \ell r = ℓ .
Magnitude of F ⃗ 13 \vv{F}_{13} F 13 (repulsive: q 1 , q 3 q_1,q_3 q 1 , q 3 both positive):
F 13 = k q 1 q 3 ℓ 2 = 8,99 ⋅ 10 9 ⋅ ( 2 ⋅ 10 − 4 ) ( 5 ⋅ 10 − 4 ) ( 0,1 ) 2 = 8,99 ⋅ 10 9 ⋅ 10 − 7 10 − 2 = 8,99 ⋅ 10 4 N ≈ 89,9 kN F_{13} = k\,\frac{q_1 q_3}{\ell^2} = 8{,}99\cdot 10^9 \cdot \frac{(2\cdot 10^{-4})(5\cdot 10^{-4})}{(0{,}1)^2} = 8{,}99\cdot 10^9 \cdot \frac{10^{-7}}{10^{-2}} = 8{,}99\cdot 10^4\ \text{N} \approx 89{,}9\ \text{kN} F 13 = k ℓ 2 q 1 q 3 = 8 , 99 ⋅ 1 0 9 ⋅ ( 0 , 1 ) 2 ( 2 ⋅ 1 0 − 4 ) ( 5 ⋅ 1 0 − 4 ) = 8 , 99 ⋅ 1 0 9 ⋅ 1 0 − 2 1 0 − 7 = 8 , 99 ⋅ 1 0 4 N ≈ 89 , 9 kN
Magnitude of F ⃗ 23 \vv{F}_{23} F 23 (attractive: q 2 < 0 q_2<0 q 2 < 0 , q 3 > 0 q_3>0 q 3 > 0 ):
F 23 = k ∣ q 2 ∣ q 3 ℓ 2 = 8,99 ⋅ 10 9 ⋅ ( 3 ⋅ 10 − 4 ) ( 5 ⋅ 10 − 4 ) ( 0,1 ) 2 = 8,99 ⋅ 10 9 ⋅ 1,5 ⋅ 10 − 7 10 − 2 ≈ 134,9 kN F_{23} = k\,\frac{|q_2|\,q_3}{\ell^2} = 8{,}99\cdot 10^9 \cdot \frac{(3\cdot 10^{-4})(5\cdot 10^{-4})}{(0{,}1)^2} = 8{,}99\cdot 10^9 \cdot \frac{1{,}5\cdot 10^{-7}}{10^{-2}} \approx 134{,}9\ \text{kN} F 23 = k ℓ 2 ∣ q 2 ∣ q 3 = 8 , 99 ⋅ 1 0 9 ⋅ ( 0 , 1 ) 2 ( 3 ⋅ 1 0 − 4 ) ( 5 ⋅ 1 0 − 4 ) = 8 , 99 ⋅ 1 0 9 ⋅ 1 0 − 2 1 , 5 ⋅ 1 0 − 7 ≈ 134 , 9 kN
Directions. Place q 3 q_3 q 3 at the top vertex (C). F ⃗ 13 \vv{F}_{13} F 13 is repulsive, so it points along the line A → C A\to C A → C extended, i.e. in direction ( cos 60 ∘ , sin 60 ∘ ) = ( 0,5 ; 0,866 ) (\cos 60^\circ,\ \sin 60^\circ) = (0{,}5;\ 0{,}866) ( cos 6 0 ∘ , sin 6 0 ∘ ) = ( 0 , 5 ; 0 , 866 ) . F ⃗ 23 \vv{F}_{23} F 23 is attractive, so it points from C C C towards B B B , direction ( 0,5 ; − 0,866 ) (0{,}5;\ -0{,}866) ( 0 , 5 ; − 0 , 866 ) .
Components (in kN):
F x = F 13 ⋅ 0,5 + F 23 ⋅ 0,5 = 89,9 ⋅ 0,5 + 134,9 ⋅ 0,5 = 44,95 + 67,45 = 112,4 kN F_x = F_{13}\cdot 0{,}5 + F_{23}\cdot 0{,}5 = 89{,}9\cdot 0{,}5 + 134{,}9\cdot 0{,}5 = 44{,}95 + 67{,}45 = 112{,}4\ \text{kN} F x = F 13 ⋅ 0 , 5 + F 23 ⋅ 0 , 5 = 89 , 9 ⋅ 0 , 5 + 134 , 9 ⋅ 0 , 5 = 44 , 95 + 67 , 45 = 112 , 4 kN
F y = F 13 ⋅ 0,866 − F 23 ⋅ 0,866 = ( 89,9 − 134,9 ) ⋅ 0,866 = − 45 ⋅ 0,866 = − 38,96 kN F_y = F_{13}\cdot 0{,}866 - F_{23}\cdot 0{,}866 = (89{,}9 - 134{,}9)\cdot 0{,}866 = -45\cdot 0{,}866 = -38{,}96\ \text{kN} F y = F 13 ⋅ 0 , 866 − F 23 ⋅ 0 , 866 = ( 89 , 9 − 134 , 9 ) ⋅ 0 , 866 = − 45 ⋅ 0 , 866 = − 38 , 96 kN
Resultant magnitude:
F = F x 2 + F y 2 = 112,4 2 + 38,96 2 = 12634 + 1518 ≈ 118,9 kN F = \sqrt{F_x^2 + F_y^2} = \sqrt{112{,}4^2 + 38{,}96^2} = \sqrt{12634 + 1518} \approx 118{,}9\ \text{kN} F = F x 2 + F y 2 = 112 , 4 2 + 38 , 9 6 2 = 12634 + 1518 ≈ 118 , 9 kN
Direction (angle below the horizontal):
θ = arctan ∣ F y ∣ F x = arctan 38,96 112,4 ≈ 19,1 ∘ \theta = \arctan\!\frac{|F_y|}{F_x} = \arctan\frac{38{,}96}{112{,}4} \approx 19{,}1^\circ θ = arctan F x ∣ F y ∣ = arctan 112 , 4 38 , 96 ≈ 19 , 1 ∘
F ≈ 118,9 kN , θ ≈ 19,1 ∘ below the horizontal \ev{F \approx 118{,}9\ \text{kN},\quad \theta \approx 19{,}1^\circ \text{ below the horizontal}} F ≈ 118 , 9 kN , θ ≈ 19 , 1 ∘ below the horizontal