## Charge in conductors

1. The problem statement, all variables and given/known data
A sphere conductor of radius 18 cm has potential 27 Volt. Another sphere conductor has potential 18 Volt. Both of them are connected and the combined potential is 24 Volt. Find:
a. the charge of second sphere
b. the charge of each sphere now

2. Relevant equations
Q = CV
V = kQ / r

3. The attempt at a solution
a.
V1 = k.Q1 / r1
27 = 9 x 109 x Q1 / (18 x 10-2)
Q1 = 5.4 x 10-10 C

Then I don't know how to continue.....

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 Try using conservation of charge. EDIT: Are those two sphere of same radius?
 Recognitions: Homework Help You can calculate also the final charge Q1' on the first sphere, as its potential is known. As for the second sphere, kQ2/R2=18 and kQ'2/R2=24. From here, you can find the ratio Q2'/Q2. As AGNuke said, the sum of the charge on the spheres is conserved: Q1+Q2=Q1'+Q2'. ehild

## Charge in conductors

 Quote by AGNuke Try using conservation of charge. EDIT: Are those two sphere of same radius?
I don't know but maybe they are not

 Quote by ehild You can calculate also the final charge Q1' on the first sphere, as its potential is known. As for the second sphere, kQ2/R2=18 and kQ'2/R2=24. From here, you can find the ratio Q2'/Q2. As AGNuke said, the sum of the charge on the spheres is conserved: Q1+Q2=Q1'+Q2'. ehild
V' = k Q1' / r1
24 = 9 x 109 x Q1' / (18 x 10-2)
Q1' = 4.8 x 10-10 C

kQ2/R2=18 ; kQ'2/R2=24
So Q2'/Q2 = 24/18 = 4/3

Q1+Q2=Q1'+Q2'
5.4 x 10-10 + Q2 = 4.8 x 10-10 + 4/3 Q2
Q2 = 1.8 x 10-10 C

Q2' = 2.4 x 10-10 C

If the spheres have same radius, then the combined potential should be: (27 + 18)/2 = 45/2 V. Am I correct in this case?

Thanks

Recognitions:
Homework Help
 Quote by songoku I don't know but maybe they are not Q2 = 1.8 x 10-10 C Q2' = 2.4 x 10-10 C If the spheres have same radius, then the combined potential should be: (27 + 18)/2 = 45/2 V. Am I correct in this case? Thanks
Your solution is excellent and you are right, if the radii were the same the final voltage would be 22.5 V.

ehild

 OK. Thanks a lot