Question

The drawing shows an equilateral triangle, each side of which has a length of 3.89 cm. Point charges are fixed to each corner, as shown. The 4.00 C charge experiences a net force due to the charges qA and qB. This net force points vertically downward and has a magnitude of 256 N. Determine (a) charge qA, (b) charge qB.

Answer 1

qA and qB are same since the net force is pointing vertically down. If qA were greater than qB, the net force would have a component in the direction of qA.
qA = qB = q
Force between q and 4 C charge, F = (k 4 q)/r² = [(8.99 x 10⁹) x 4q] / (r)².
r = 0.0389 m.
Net force = 2F sin(60) [The horizontal components will cancel each other when we take the vector sum of the two forces]
Net force = 2 x [(8.99 x 10⁹) x 4q] / (0.0389)².x sin(60) = 256 N
q x 10⁹ = 6.22 x 10^-3.
q = 6.22 x 10^-12 C