Charge and energy of a conducting sphere

In summary, a conducting sphere with a radius of 100cm is charged to a potential of 30 volts. The charge on the sphere is 3.3X10^-7 coulombs and the energy stored in the field is 4.95X10^-6 joules. When the sphere is connected to a second identical uncharged sphere by a long wire, the final energy in the system can be calculated by considering the transfer of charge between the two spheres, taking into account electrical potential and electric currents. The final energy will not necessarily be double the initial energy as the final charge on each sphere will not be the same as the initial charge on the single sphere.
  • #1
AsadaShino92
21
0

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.
 
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  • #2
The wire connecting the two spheres is "long", so you can treat each sphere as being isolated.
 
  • #3
So since each sphere has the same charge does that mean the final energy will double?
 
  • #4
AsadaShino92 said:

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?
 
  • #5
AsadaShino92 said:
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).
 
  • #6
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.
 

Related to Charge and energy of a conducting sphere

What is the charge on a conducting sphere?

The charge on a conducting sphere is the sum of all the charges (positive and negative) present on the surface of the sphere. This charge is distributed evenly on the surface of the sphere due to the conducting properties of the material.

How is the charge on a conducting sphere distributed?

The charge on a conducting sphere is distributed evenly on the surface of the sphere. This is known as the "Faraday's Law of Electrostatic Induction". The charge distributes itself in a way that the electric field inside the sphere is always zero.

Can the charge on a conducting sphere be changed?

Yes, the charge on a conducting sphere can be changed by bringing it in contact with another charged object or by applying an external electric field. The charge on the sphere will redistribute itself accordingly, but the total charge will remain the same.

What is the energy of a conducting sphere?

The energy of a conducting sphere is a measure of the work required to assemble all the charges on the surface of the sphere. This energy is directly proportional to the square of the charge and inversely proportional to the radius of the sphere.

How does the energy of a conducting sphere change with distance?

The energy of a conducting sphere decreases as the distance from the center of the sphere increases. This is because the electric field strength also decreases with distance. As the charges are distributed evenly on the surface, the energy is spread out over a larger area, resulting in a decrease in energy.

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