Question

A charge, q₁, of +16 micro-coulombs is located at x = 0, y = 6.4 cm, a charge q₂ of -40 micro-coulombs is located at x = 0, y = -2.8 cm, an unknown charge of q₃, is located at x =- 3.3 cm, y = 0 cm, and a charge, q₄, -6 micro-coulombs is located at x = +16.6 cm, y = 0 cm. The direction of the total electric force on q₄ is 136.5 degrees from the +x axis, counter-clockwise. What is the unknown charge, q₃, in micro-coulombs? If negative, include a negative sign in your answer.

Please explain what you are doing!

Answer 1

16 x 10 C 0.064 m F2 0.033 m 0.166 m 0 Q3-? .028m 4-6 x 104 c 89 40 x 10c °С

F = Net force on charge q₄

F_x = x-component of the net force = F Cos136.5 = - 0.725 F

F_y = y-component of the net force = F Sin136.5 = 0.688 F

$\theta _{1}$ = tan^-1(0.064/0.166) = 21.1 deg

$\theta _{2}$ = tan^-1(0.028/0.166) = 9.6 deg

r₁ = distance of charge q₁ from q₄ = sqrt(OA² + OD²) = sqrt((0.064)² + (0.166)²) = 0.178 m

r₂ = distance of charge q₂ from q₄ = sqrt(OB² + OD²) = sqrt((0.028)² + (0.166)²) = 0.168 m

r₃ = distance of charge q₃ from q₄ = 0.033 + 0.166 = 0.199 m

F₁ = force by charge q₁ on q₄ = k q₁ q₄/r₁² = (9 x 10⁹) (16 x 10^-6) (6 x 10^-6)/(0.178)² = 27.3 N

F₂ = force by charge q₂ on q₄ = k q₂ q₄/r₁² = (9 x 10⁹) (40 x 10^-6) (6 x 10^-6)/(0.168)² = 76.5 N

F₃ = force by charge q₃ on q₄ = k q₃ q₄/r₃² = (9 x 10⁹) q₃ (6 x 10^-6)/(0.199)² = (1.4 x 10⁶) q₃

Net force along the Y-direction is given as

F_y = F₁ Sin$\theta _{1}$ + F₂ Sin$\theta _{2}$

F_y = (27.3) Sin21.1 + (76.5) Sin9.6

F_y = 22.6

0.688 F = 22.6

F = 32.85 N

force along the x-direction is given as

F_x = - F₁ Cos$\theta _{1}$ + F₂ Cos$\theta _{2}$ - F₃

- 0.725 F = - F₁ Cos$\theta _{1}$ + F₂ Cos$\theta _{2}$ + F₃

- (0.725) (32.85) = - (27.3) Cos21.1 + (76.5) Cos9.6 - (1.4 x 10⁶) q₃

q₃ = 52.7 x 10^-6 C