Question

The charges q, 3.0 10 C -6.5 10C, and1.910Careed at the cornersof +y-axis is up along the page.) 5.0 m 3.0 m 4.0 m 93

Answer 1

Force on q₁ due to charge q₂ which will be given as -

F₂₁ = k_e |q₂| |q₁| / r₂₁²

F₂₁ = [(9 x 10⁹ Nm²/C²) (6.5 x 10^-7 C) (3.8 x 10^-7 C)] / (5 m)²

F₂₁ = 8.89 x 10^-5 N

Force on q₁ due to charge q₃ which will be given as -

F₃₁ = k_e |q₃| |q₁| / r₃₁²

F₃₁ = [(9 x 10⁹ Nm²/C²) (1.9 x 10^-7 C) (3.8 x 10^-7 C)] / (3 m)²

F₃₁ = 7.22 x 10^-5 N

The x-component of force on q₁ which will be given as -

F_1,x = F_21,x + F_31,x

F_1,x = (8.89 x 10^-5 N) cos 60⁰ + (7.22 x 10^-5 N) cos 90⁰

F_1,x = (8.89 x 10^-5 N) (0.5)

F_1,x = 4.445 x 10^-5 N

The y-component of force on q₁ which will be given as -

F_1,y = F_21,y + F_31,y

F_1,y = (8.89 x 10^-5 N) sin 60⁰ + (7.22 x 10^-5 N) sin 90⁰

F_1,y = [(8.89 x 10^-5 N) (0.866) + (7.22 x 10^-5 N)]

F_1,y = 1.491 x 10^-4 N

Therefore, the force on charge q₁ which will be given as -

F_{on 1} = _$\sqrt{}$F_1,x² + F_1,y²

F_{on 1} = _$\sqrt{}$(4.445 x 10^-5 N)² + (1.491 x 10^-4 N)²

F_{on 1} = _$\sqrt{}$2.420 x 10^-8 N²

F_{on 1} = 1.55 x 10^-4 N