# Charge and energy of a conducting sphere

1. Dec 10, 2015

1. The problem statement, all variables and given/known data

A conducting sphere of radius 100cm is charged to a potential of 30 volts.
a) What is the charge on the sphere?
b) What is the energy stored in the field?
c) If the sphere is connected to a second identical uncharged sphere by a long wire, what is the final energy in the system? You can neglect any interference.

2. Relevant equations

Q=CV
C=4πεr
U=(1/2)CV^2

3. The attempt at a solution

For part A I find the capacitance of the sphere so I can then apply Q=CV to find the charge.
C=4π(8.85X10^-12)(100)=1.1X10^-8

Q=(1.1X10^-8)(30)=3.3X10^-7

For part B I used the energy equation U=(1/2)(1.1X10^-8)(900)=4.95X10^-6

I am pretty much stuck on part C. Since it is a conducting sphere the potential for both of them should be the same. So I tried equating both potential equations such as V=(kq1)/(r)=(kq2)/(r). Both spheres are identical so I get that q1=q2. Not really clear on how to find the final energy.

2. Dec 10, 2015

### TSny

The wire connecting the two spheres is "long", so you can treat each sphere as being isolated.

3. Dec 10, 2015

So since each sphere has the same charge does that mean the final energy will double?

4. Dec 10, 2015

### Staff: Mentor

What are the units? I see you've used "100" for the radius. 100 what? So what are the capacitance units you end up with?
Again, what are the units for these?

5. Dec 10, 2015

### TSny

No. The final charge on each sphere is not the same as the initial charge on the one sphere (before the wire was connected).

6. Dec 10, 2015

### J Hann

If the initial sphere has charge Q and then you connect the spheres by a long wire then
charge q will be transferred to the uncharged sphere.
Now using what you know about electrical potential and electric currents
you should be able to calculate the final charge configuration.