Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 47.8 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0720 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.115 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1

Answer 1

Electrostatic force is given by:

F = kq1q2/d^2

d = 47.8 cm = 0.478 m

F1 = 0.0720 N

q1*q2 = F*d^2/k

q1*q2 = 0.0720*0.478^2/(9*10^9) = 1.83*10^-12

Now when both spheres are connected with wire

Charge on each sphere will be

q = (q1 + q2)/2

F2 = k*q*q/d^2

q = sqrt (F2*d^2/k)

q = sqrt (0.115*0.478^2/(9*10^9))

q = 1.71*10^-6 C

(q1 + q2)/2 = 1.71*10^-6

q1 + q2 = 3.42*10^-6

q1*q2 = 1.83*10^-12

(3.42*10^-6 - q2)*q2 = 1.83*10^-12

Solving above equation:

q2 = 2.76*10^-6 C

q1 = 0.66*10^-6 C