Question

Chapter 18, Problem 20 The drawing shows an equilateral triangle, each side of which has a length of 1.42 cm. Point charges are fixed to each corner, as shown. The 4.00 μC charge experiences a net force due to the charges qA and qB. This net force points vertically downward and has a magnitude of 708 N Determine (a) charge qA, (b) charge qB 4.00 IA TB

Answer 1

Since the net force is vertically downward then both qA and qB are negative and have equal magnitude

So let qA = qB = q

Now F = k*q1*qA/r^2*cos(30) + k*q1*qB*cos(30)/r^2 = 2*k*q1*q/r^2*cos(30)

so q = (F*r^2/(2*k*q1*cos(30)) = (708*(1.42x10^-2)^2/(2*9.0x10^9*4.00x10^-6*cos30) = 2.29x10^-6C

(a) So qA = -2.29x10^-6 C

(b) and qB = -2.29x10^-6 C