# Charge in conductors

by songoku
Tags: charge, conductors
 P: 770 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.....
 PF Gold P: 456 Try using conservation of charge. EDIT: Are those two sphere of same radius?
 HW Helper Thanks P: 10,759 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
P: 770
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
HW Helper
Thanks
P: 10,759
 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
 P: 770 OK. Thanks a lot

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