Charge and energy of a conducting sphere

[AsadaShino92]

Homework Statement

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.

Homework Equations

Q=CV
C=4πεr
U=(1/2)CV^2

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.

 

[TSny]

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

 

[AsadaShino92]

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

 

[gneill]

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

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?

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

Again, what are the units for these?

 

[TSny]

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

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

 

[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.