Question

The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +7.80 µC; the other two charges have identical magnitudes, but opposite signs: q2 = -4.85 µC and q3 = +4.85 µC.
a.   Determine the net force exerted on q2 by the other two charges.
b.   If q2 had a mass of 1.62 g and it were free to move, what would be its acceleration?

45 FCB 45), FBC 2 FAB l OT FAC

Answer 1

In triangle ABC , using Pythagorean theorem

AC = sqrt(AB² + BC²) = sqrt((10^-2)² + (10^-2)²) = 1.41 x 10^-2 m

F₁ = force by charge q₁ on q₂ = k q₁ q₂/r₁² = (9 x 10⁹) (7.80 x 10^-6) (4.85 x 10^-6)/(10^-2)² = 3404.7 N

F₃ = force by charge q₃ on q₂ = k q₃ q₂/r₁² = (9 x 10⁹) (4.85 x 10^-6) (4.85 x 10^-6)/(1.41 x 10^-2)² = 1064.85 N

F_3x = X-component of F₃ = F₃ Cos45 = 1064.85 Cos45 = 753 N

F_3y = Y-component of F₃ = F₃ Sin45 = 1064.85 Sin45 = 753 N

Net force along the X-direction is given as

F_x = F_3x = 753 N

Net force along the Y-direction is given as

F_y = - F_3y - F₁ = - 753 - 3404.7 = - 4157.7 N

Net force on charge q2 is given as

F_net = sqrt(F_x² + F_y²)

F_net = sqrt((753)² + (- 4157.7)²) = 4225.34 N

direction : $\theta$ = tan^-1(F_y/F_x) = tan^-1(-4157.7/753) = - 79.73 deg   

negative sign indicate direction below positive x-direction

b)

m = mass of charge q₂ = 1.62 g = 1.62 x 10^-3 kg

acceleration is given as

a = F_net/m

a = 4225.34/(1.62 x 10^-3)

a = 2.61 x 10⁶ m/s²