Problem Charges 2 β ΞΌ C 2 \, \mu \mathrm{C} 2 ΞΌ C β 3 β ΞΌ C -3 \, \mu \mathrm{C} β 3 ΞΌ C 4 β ΞΌ C 4 \, \mu \mathrm{C} 4 ΞΌ C 10 β c m 10 \, \mathrm{cm} 10 cm 4 β ΞΌ C 4 \, \mu \mathrm{C} 4 ΞΌ C
Data
q 1 = 2 β ΞΌ C = 2 Γ 10 β 6 β C q_1 = 2 \, \mu \mathrm{C} = 2 \times 10^{-6} \, \mathrm{C} q 1 β = 2 ΞΌ C = 2 Γ 1 0 β 6 C q 2 = β 3 β ΞΌ C = β 3 Γ 10 β 6 β C q_2 = -3 \, \mu \mathrm{C} = -3 \times 10^{-6} \, \mathrm{C} q 2 β = β 3 ΞΌ C = β 3 Γ 1 0 β 6 C q 3 = 4 β ΞΌ C = 4 Γ 10 β 6 β C q_3 = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C} q 3 β = 4 ΞΌ C = 4 Γ 1 0 β 6 C Side length, r = 10 β c m = 0.1 β m r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m} r = 10 cm = 0.1 m
Coulombβs constant, k = 9 Γ 10 9 β N m 2 C β 2 k = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}} k = 9 Γ 1 0 9 N m 2 C β 2
Prerequisite Concepts
Coulombβs Force Formula :
F = k q 1 q 2 r 2 F = \frac{k q_1 q_2}{r^2} F = r 2 k q 1 β q 2 β β Answer Step 1: Force due to q 1 q_1 q 1 β
F 1 = k q 1 q 3 r 2 F_1 = \frac{k q_1 q_3}{r^2} F 1 β = r 2 k q 1 β q 3 β β Substitute values:
F 1 = ( 9 Γ 10 9 ) ( 2 Γ 10 β 6 ) ( 4 Γ 10 β 6 ) ( 0.1 ) 2 F_1 = \frac{(9 \times 10^9)(2 \times 10^{-6})(4 \times 10^{-6})}{(0.1)^2} F 1 β = ( 0.1 ) 2 ( 9 Γ 1 0 9 ) ( 2 Γ 1 0 β 6 ) ( 4 Γ 1 0 β 6 ) β F 1 = 7.2 β N F_1 = 7.2 \, \mathrm{N} F 1 β = 7.2 N Step 2: Force due to q 2 q_2 q 2 β
F 2 = k q 2 q 3 r 2 F_2 = \frac{k q_2 q_3}{r^2} F 2 β = r 2 k q 2 β q 3 β β Substitute values:
F 2 = ( 9 Γ 10 9 ) ( 3 Γ 10 β 6 ) ( 4 Γ 10 β 6 ) ( 0.1 ) 2 F_2 = \frac{(9 \times 10^9)(3 \times 10^{-6})(4 \times 10^{-6})}{(0.1)^2} F 2 β = ( 0.1 ) 2 ( 9 Γ 1 0 9 ) ( 3 Γ 1 0 β 6 ) ( 4 Γ 1 0 β 6 ) β F 2 = 10.8 β N F_2 = 10.8 \, \mathrm{N} F 2 β = 10.8 N Step 3: Resultant Force Components
The triangle is equilateral, so each force makes an angle of 60 β 60^\circ 6 0 β
X-component:
F x = F 1 cos β‘ 60 β + F 2 cos β‘ 60 β F_x = F_1 \cos 60^\circ + F_2 \cos 60^\circ F x β = F 1 β cos 6 0 β + F 2 β cos 6 0 β F x = 7.2 Γ 0.5 + 10.8 Γ 0.5 = 9 β N F_x = 7.2 \times 0.5 + 10.8 \times 0.5 = 9 \, \mathrm{N} F x β = 7.2 Γ 0.5 + 10.8 Γ 0.5 = 9 N Y-component:
F y = F 1 sin β‘ 60 β β F 2 sin β‘ 60 β F_y = F_1 \sin 60^\circ - F_2 \sin 60^\circ F y β = F 1 β sin 6 0 β β F 2 β sin 6 0 β F y = 7.2 Γ 0.866 β 10.8 Γ 0.866 = β 3.2 β N F_y = 7.2 \times 0.866 - 10.8 \times 0.866 = -3.2 \, \mathrm{N} F y β = 7.2 Γ 0.866 β 10.8 Γ 0.866 = β 3.2 N Step 4: Magnitude of Resultant Force
F = F x 2 + F y 2 F = \sqrt{F_x^2 + F_y^2} F = F x 2 β + F y 2 β β F = 9 2 + ( β 3.2 ) 2 = 91.24 = 9.55 β N F = \sqrt{9^2 + (-3.2)^2} = \sqrt{91.24} = 9.55 \, \mathrm{N} F = 9 2 + ( β 3.2 ) 2 β = 91.24 β = 9.55 N Step 5: Direction of Resultant Force
ΞΈ = tan β‘ β 1 ( F y F x ) \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) ΞΈ = tan β 1 ( F x β F y β β ) ΞΈ = tan β‘ β 1 ( β 3.2 9 ) = β 19.5 β \theta = \tan^{-1} \left( \frac{-3.2}{9} \right) = -19.5^\circ ΞΈ = tan β 1 ( 9 β 3.2 β ) = β 19. 5 β Final Answer:
Magnitude: 9.55 β N 9.55 \, \mathrm{N} 9.55 N
Direction: β 19.5 β -19.5^\circ β 19. 5 β
