Question

Sun 6:11 PM a 66% E- uni The charges and coordinates of two charged partides held fixed in an xy plane are q1 " 3.02 C, x1 3.56 cm, yi-0.970 cm and and (b) direction (with respect to +x-axis in the range (-180°;1801) of the electrostatic force on should a third particle of charge 93-3.53 uC be placed such that the net electrostatic force on particle 2 2 due to particle 1. At what (c) x and (d) y coordinates due to particles 1 and 3 is zero? (a) Number (b) Number (c) Number (d) Number

Answer 1

-2.62 cm, 2.48 cm 21 (3.56 cm, 0.97 cm) 0

Distance from particle 2 due to particle 1 which will be given as -

r₂₁ = _$\sqrt{}$(x₂ - x₁)² + (y₂ - y₁)²

r₂₁ = _$\sqrt{}$[(-0.0262 m) - (0.0356 m)]² + [(0.0248 m) - (0.0097 m)]²

r₂₁ = _$\sqrt{}$0.00404725 m²

r₂₁ = 0.063 m

(a) Magnitude of an electrostatic force on particle 2 due to particle 1 which will be given by -

| F₂₁ | = k_e |q₁| |q₂| / r₂₁²

| F₂₁ | = [(9 x 10⁹ Nm²/C²) (3.02 x 10^-6 C) (6.20 x 10^-6 C)] / (0.00404725 m²)

| F₂₁ | = 41.6 N

(b) Direction of an electrostatic force on particle 2 due to particle 1 which will be given by -

tan $\theta$ = [(y₂ - y₁) / (x₂ - x₁)]

tan $\theta$ = { [(0.0248 m) - (0.0097 m)] / [(-0.0262 m) - (0.0356 m)] }

$\theta$ = tan^-1 (-0.2443)

$\theta$ = - 13.7 degree

{ below the positive x-axis }

We know that, | F₂₃ | = | F₂₁ |

k_e |q₂| |q₃| / r₂₃² = k_e |q₁| |q₂| / r₂₁²

r₂₃² = r₂₁² |q₃| / |q₁|

r₂₃ = _$\sqrt{}$[(0.00404725 m) (3.53 x 10^-6 C)] / (3.02 x 10^-6 C)

r₂₃ = _$\sqrt{}$0.00473 m²

r₂₃ = 0.0687 m

converting m into cm :

r₂₃ = 6.87 cm

(c) At what x coordinates should a third particle of charge q₃ be placed?

x₃ = x₂ - r₂₃ cos $\theta$

x₃ = (-2.62 cm) - (6.87 cm) cos 13.7⁰

x₃ = [(-2.62 cm) - (6.67 cm)]

x₃ = - 9.29 cm

(d) At what y coordinates should a third particle of charge q₃ be placed?

y₃ = y₂ - r₂₃ sin $\theta$

y₃ = (2.48 cm) - (6.87 cm) sin 13.7⁰

y₃ = [(2.48 cm) + (1.62 cm)]

y₃ = 4.10 cm