Question

Please help me to solve the question #5 with an explanation:

Three charges (q1 = 5.8 pC q2 =-5.9 pC and q3-3.4pC are located at the vertices of an equilateral triangle with side d 9.3 cm as shown. yt 43 91 1) What is F3,x the value of the x-component of the net force on q3? 20.70 Submit 2) What is F3,y, the value of the y-component of the net force on q3? 0.3064 Submit

Thank you!

Answer 1

given

q₁ = 5.8 uC

q₂ = - 5.9 uC

q₃ = 3.4 uC

d = 9.3 cm = 0.093 m

cos $\theta$ = d/2 / d

$\theta$ = 60^o

a )

F_3x = k q₃ ( q₁ + q₂ ) cos $\theta$/ d²

= 9 x 10⁹ x 3.4 x 10^-6 x ( 5.8 + 5.9 ) x 10^-6 x cos60 / 0.093²

F_3x = 20.69 N

b )

F_3y = k q₃ ( q₁ - q₂ ) sin $\theta$/ d²

= 9 x 10⁹ x 3.4 x 10^-6 x ( 5.8 -0.0 5.9 ) x 10^-6 x sin60 / 0.093²

F_3y = - 0.306 N

c )

F_2x = - ( k q₂ ( q₁ + q₂ + q₃ + q₄ ) cos $\theta$/ d²

= - 9 x 10⁹ x 5.9 x 10^-6 x cos60 x ( 5.8 + 5.9 + 3.4 + 5.9 ) x 10^-6 / 0.093²

F_2x = - 64.46 N

d )

F_2x = k q₂ ( q₃ - q₄ ) sin $\theta$/ d²

= 9 x 10⁹ x 5.9 x 10^-6 x sin60 x ( 0 ) x 10^-6 / 0.093²

F_2x = 0 N

e )

F_1x = k q₁ ( q₂ - q₃ - q₄ ) cos $\theta$/ d²  

= 9 x 10⁹ x 5.8 x 10^-6 x cos60 x (5.9 - 3.4 - 3.4 ) x 10^-6 / 0.093²

F_1x = - 4.7 N