Question

*the answer is not 5.12e-12

Chapter 18, Problem 20 Your answer is partially correct. Try again. The drawing shows an equilateral triangle, each side of which has a length of 2.29 cm. Pcint charges are fixed to each corner, as shown. The 4.00 c charge experiences a net force due to the charges and B This net force points vertically downward and has a magnitude of 703 N. Determine (a) charge gA, (b) charge +4.00 C (a) Number T5.918e-6 (b) Number UnitśTc SHOW HINT SHOW SOLUTION LINK TO TEXT

Answer 1

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

qA = qB = Q

Now, F = kq*qA*cos 30 degree/r^2 + kq*qB*cos 30 degree /r^2

= 2*k*q*Q*cos 30 degree /r^2

Q = F*r^2 /(2*k*q*cos 30 degree)

= 703*(2.29*10^-2)^2 /(2*9*10^9*4*10^-6*cos 30 degree)

= 5.91*10^-6 C

Q = qA = qB = - 5.91*10^-6 C

Answer

and qB = -1.84x10^-6 C