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Charge in conductors 
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#1
Aug1312, 07:01 AM

P: 767

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. V_{1} = k.Q_{1} / r_{1} 27 = 9 x 10^{9} x Q_{1} / (18 x 10^{2}) Q_{1} = 5.4 x 10^{10} C Then I don't know how to continue..... 


#2
Aug1312, 07:09 AM

PF Gold
P: 456

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


#3
Aug1312, 07:52 AM

HW Helper
Thanks
P: 10,380

You can calculate also the final charge Q1' on the first sphere, as its potential is known.
As for the second sphere, kQ_{2}/R_{2}=18 and kQ'_{2}/R_{2}=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 


#4
Aug1312, 11:22 PM

P: 767

Charge in conductors
24 = 9 x 10^{9} x Q1' / (18 x 10^{2}) Q1' = 4.8 x 10^{10} C kQ_{2}/R_{2}=18 ; kQ'_{2}/R_{2}=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 


#5
Aug1312, 11:37 PM

HW Helper
Thanks
P: 10,380

ehild 


#6
Aug1412, 04:26 AM

P: 767

OK. Thanks a lot



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