Question

What is the magnitude and direction (relative to the +a axis) of the net electric force due to the other three charges on qi in the upper-left corner of the rectangle below? +3 42 q1 91 = +1 C 5 m 3 m 43 = 36.90 4 m

Answer 1

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

F₁₃ = k_e |q₁| |q₃| / r₁₃²

F₁₃ = [(9 x 10⁹ Nm²/C²) (1 C) (3 C)] / (5 m)²

F₁₃ = 1.08 x 10⁹ N

The x and y components of F₁₃ which are given below as -

F_13,x = (1.08 x 10⁹ N) cos 36.9⁰

F_13,x = 8.63 x 10⁸ N

And

F_13,y = (1.08 x 10⁹ N) sin 36.9⁰

F_13,y = 6.48 x 10⁸ N

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

F₁₂ = k_e |q₁| |q₂| / r₁₂²

F₁₂ = [(9 x 10⁹ Nm²/C²) (1 C) (2 C)] / (4 m)²

F₁₂ = 1.125 x 10⁹ N

The x and y components of F₁₂ which are given below as -

F_12,x = (1.125 x 10⁹ N) cos 0⁰

F_12,x = 1.125 x 10⁹ N

And

F_12,y = (1.125 x 10⁹ N) sin 0⁰

F_12,y = 0 N

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

F₁ = (F_13,x + F_12,x) + (F_13,y + F_12,y)

F₁ = [(8.63 x 10⁸ N) + (1.125 x 10⁹ N)] + [(6.48 x 10⁸ N) + (0 N)]

F₁ = (1.98 x 10⁹ N) + (6.48 x 10⁸ N)

Therefore, magnitude of the net electric force due to other three charges on q₁ which will be given by -

| F₁ | = _$\sqrt{}$(1.98 x 10⁹ N)² + (6.48 x 10⁸ N)²

| F₁ | = _$\sqrt{}$4.34 x 10¹⁸ N²

| F₁ | = 2.08 x 10⁹ N

From a trigonometry identity, we have

_$\phi$ = tan^-1 [(6.48 x 10⁸ N) / (1.98 x 10⁹ N)]

_$\phi$ = tan^-1 (0.3272)

_$\phi$ = 18.1 degree