Question

A charge, q1, of -2 micro-coulombs is located at the origin, a charge, q2, of -8 micro-coulombs is located at x = 0 cm, y = +5.5 cm, a charge, q3, of +3 micro-coulombs is located at x = 0 cm, y = -5.2 cm, and a charge, q4, of +12 micro-coulombs is located at x = 2.2 cm, y = 0 cm. What is the angle of the total electric force on q2 (charge on the +y axis) in degrees measured counter-clockwise from the +x axis?

Answer 1

Using a trigonometric identity, we have

$\theta$ = tan^-1 [(5.5 cm) / (2.2 cm)]

$\theta$ = tan^-1 (2.5)

$\theta$ = 68.2 degree

The force on charge q₂ due to q₄ which will be given as -

F₂₄ = k_e |q₂| |q₄| / r₂₄²

F₂₄ = [(9 x 10⁹ Nm²/C²) (8 x 10^-6 C) (12 x 10^-6 C)] / (0.0592 m)²

F₂₄ = 246.5 N

The x and y components of F₂₄ which are given below as -

F_24,x = (246.5 N) cos 68.2⁰

F_24,x = 91.5 N

And

F_24,y = (246.5 N) sin 68.2⁰

F_24,y = 228.8 N

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

F₂₁ = k_e |q₂| |q₁| / r₂₁²

F₂₁ = [(9 x 10⁹ Nm²/C²) (8 x 10^-6 C) (2 x 10^-6 C)] / (0.055 m)²

F₂₁ = 47.6 N

The x and y components of F₂₁ which are given below as -

F_21,x = (47.6 N) cos 90⁰

F_21,x = 0 N

And

F_21,y = (47.6 N) sin 90⁰

F_21,y = 47.6 N

The net electric force due to other three charges on q₁ which will be given as -

F₂ = (F_24,x + F_21,x) + (F_24,y + F_21,y)

F₂ = [(91.5 N) + (0 N)] + [(228.8 N) + (47.6 N)]

F₂ = (91.5 N) + (276.4 N)

Therefore, magnitude of the net electric force due to other three charges on q₂ which will be given by -

| F₂ | = _$\sqrt{}$(91.5 N)² + (276.4 N)²

| F₂ | = _$\sqrt{}$84769.21 N²

| F₂ | = 291.1 N

The angle of total electric force on charge q₂ which will be given as -

From a trigonometry identity, we get

_$\phi$ = tan^-1 [(276.4 N) / (91.5 N)]

_$\phi$ = tan^-1 (3.0207)

_$\phi$ = 71.6 degree

{ counter-clockwise from the +x axis }