Solving Charge in Conductors: Find Q1, Q2 & Combined Potential

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Homework Statement


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


Homework Equations


Q = CV
V = kQ / r

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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Discussion

Try using conservation of charge.

EDIT: Are those two sphere of same radius?
 
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
 
AGNuke said:
Try using conservation of charge.

EDIT: Are those two sphere of same radius?

I don't know but maybe they are not

ehild said:
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
 
songoku said:
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