Question

Chapter 21, Problem 062 In the figure what are the (a) magnitude and (b) direction (from x-axis in the counterclockwise direction) of the net electrostatic force on particle 4 due to the other three particles? All four particles are fixed in the xy plane, and 41 = -14.40 x 10-19 C, 92 = +14.40 x 10-19 C, 43 = +28.80 x 10-19 C, 94 = +14.40 x 10-19 C, B, = 43°, dy = 3.90 cm, and d2 = dz -2.60 cm.

Answer 1

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The net "x" component of the force on particle 4 due to other three particles which will be given as -

F_net,x = F_42,x + F_41,x + F_43,x

F_net,x = {k_e |q₂| |q₄| / (d₁)²} cos 90⁰ + {k_e |q₁| |q₄| / (d₁)²} cos (180⁰ + $\theta$₁) + {k_e |q₃| |q₄| / (d₃)²} cos 180⁰

F_net,x = 0 + {k_e |q₁| |q₄| / (d₁)²} cos 223⁰ - k_e |q₃| |q₄| / (d₃)²

F_net,x = {[(9 x 10⁹ Nm²/C²) (14.4 x 10^-19 C) (14.4 x 10^-19 C) (-0.7313)] / (0.039 m)²} - {[(9 x 10⁹ Nm²/C²) (28.8 x 10^-19 C) (14.4 x 10^-19 C)] / (0.026 m)²}

F_net,x = - (8.97 x 10^-24 N) - (5.52 x 10^-23 N)

F_net,x = - 6.417 x 10^-23 N

The net "y" component of the force on particle 4 due to other three particles which will be given as -

F_net,y = F_42,y + F_41,y + F_43,y

F_net,y = - {k_e |q₂| |q₄| / (d₁)²} sin 90⁰ + {k_e |q₁| |q₄| / (d₁)²} sin (180⁰ + $\theta$₁) + {k_e |q₃| |q₄| / (d₃)²} sin 180⁰

F_net,y = - k_e |q₂| |q₄| / (d₂)² + {k_e |q₁| |q₄| / (d₁)²} sin 223⁰ + 0

F_net,y = - {[(9 x 10⁹ Nm²/C²) (14.40 x 10^-19 C) (14.4 x 10^-19 C)] / (0.026 m)²} + {[(9 x 10⁹ Nm²/C²) (14.4 x 10^-19 C) (14.4 x 10^-19 C) (-0.6819)] / (0.039 m)²}

F_net,y = - (2.76 x 10^-23 N) - (8.36 x 10^-24 N)

F_net,y = - 3.596 x 10^-23 N

(a) Magnitude of net electrostatic force on particle 4 due to other three particles which will be given by -

| F_net | = _$\sqrt{}$F_x² + F_y²

| F_net | = _$\sqrt{}$(-6.417 x 10^-23 N)² + (-3.596 x 10^-23 N)²

| F_net | = 7.35 x 10^-23 N

(b) Direction of net electrostatic force on particle 4 due to other three particles which will be given by -

$\theta$ = tan^-1 (F_y / F_x)

$\theta$ = tan^-1 [(-3.596 x 10^-23 N) / (-6.417 x 10^-23 N)]

$\theta$ = 29.2 degree

We know that, $\theta$' = 180⁰ + $\theta$

$\theta$' = 209.2 degree

{ from the +x axis in counterclockwise direction }