Two conducting spheres of radii 10 cm and 20 cm have equal c

Two conducting spheres of radii 10 cm and 20 cm have equal charges of 5 mu C and are separated very far apart If a wire is now connected between the two spheres, charge will flow from one sphere to the other until the potentials are equalized. a) What is the new charge charge distribution on the two spheres?, b) the new voltage on the spheres ? and c) the potential energy of each sphere?

Solution

a)
V1 = V2
K*q1/r1 = K*q2/r2
q1/r1 = q2/r2

Let charge on 10 cm sphere be q . then charge on 20 cm sphere = total charge - q = 5+5-q = 10-q

q1/r1 = q2/r2
q/10 = (10-q)/20
20q = 100 - 10q
30q =100
q = 10/3 micro coulomb

Charge on 10 cm sphere = 10/3 micro coulomb
Charge on 20 cm sphere = 20/3 micro coulomb

b)
V1 = V1 = K*q1/r1
=(9*10^9)*(10/3 * 10^-6)/(0.1)
= 300000 V
Answer: 300000 V

c)
for 10 cm sphere,
PE1 = q2*V
= (20/3)*10^-6 * 300000
= 2 J

for 20 cm sphere,
PE2 = q1*V
= (10/3)*10^-6 * 300000
= 1 J