Question

Please see the problem below, please be sure to read the numbers and question carefully and be sure to check for all unit conversions and double check your answer. Thank you!

Below the question is the outline of how to work the question out!

1) A charge, g1, of -8 micro-coulombs is located at the origin, a charge, q2, of -5 micro-coulombs is located at x- 0 cm, y +8.6 cm, a charge, q3, of +8 micro-coulombs is located at x = 0 cm, y =-2.6 cm, and a charge, q4, of +2 micro-coulombs is located at x-7.4 cm, y-0 cm, what is the magnitude of the total electric force on d2 (charge on the +y axis) in Newtons?

Answer 1

AD = distance between charge q₃ and q₂ = 8.6 + 2.6 = 11.2 cm = 0.112 m

AB = distance between charge q₁ and q₂ = 8.6 cm = 0.086 m

BC = distance between charge q₄ and q₂ = 7.4 cm = 0.074 m

Using Pythagorean theorem in triangle ABC

AC = sqrt(AB² + BC²) = sqrt((0.086)² + (0.074)²) = 0.1135 m

$\theta$ = tan^-1(BC/AB) = tan^-1(0.074/0.086) = 40.7 deg

F₁ = force by charge q₁ on q₂ = k q₁ q₂ /AB² = (9 x 10⁹) (8 x 10^-6) (5 x 10^-6)/(0.086)² = 48.7 N

F₃ = force by charge q₃ on q₂ = k q₃ q₂ /AD² = (9 x 10⁹) (8 x 10^-6) (5 x 10^-6)/(0.112)² = 28.7 N

F₄ = force by charge q₄ on q₂ = k q₄ q₂ /AC² = (9 x 10⁹) (2 x 10^-6) (5 x 10^-6)/(0.1135)² = 6.98 N

Net force along Y-direction is given as

F_y = F₁ - F₃ - F₄ Cos$\theta$ = 48.7 - 28.7 - (6.98) Cos40.7 = 14.71 N

Net force along X-direction is given as

F_x = F₄ Sin$\theta$ = (6.98) Sin40.7 = 4.55 N

magnitude of net force is given using Pythagorean theorem as

F = sqrt(F_x² + F_y²) = sqrt((4.55)² + (14.71)²) = 15.4 N