What is the heat produced when two charged spheres are connected by a wire?

[Gummy_Bear]

Homework Statement

A solid metal spheres of radius a surrounded by a concentric thin metal shell of radius 2a. Initially botb are having charge Q each. When the two are connected by a conducting wire , then the amount of heat produced in process will be :

Homework Equations

Heat = E2 - E1
Energy = kq²/ distance [/B]

The Attempt at a Solution

So i first tried to calculate the energy in initial condition.
I got E1 = kQ²/a + kQ²/2a
Then when the spheres are connected, then i think charge of inner sphere flows to outer sphere so E2 = k(2Q)²/ 2a
Then i put in the formula E2 -E1, but didn't get the answer
[/B]

 

[TSny]

Hello and welcome to PF!

Homework Equations

Energy = kq²/ distance

For what situation does this equation apply? What "distance" is used here?

3. The Attempt at a Solution
So i first tried to calculate the energy in initial condition.
I got E1 = kQ²/a + kQ²/2a

This is not correct. Can you explain how you got this expression for the total initial energy?

 

[Gummy_Bear]

Oh I am really sorry.Now that i saw the expression i realized that expression for initial energy is wrong .
And please can you give me some hints to solve this problem.

 

[TSny]

It helps to know your background. Can you use calculus in solving the problem? Can you relate this problem to any other problems you've seen?

Suppose you didn't have the shell, but just the sphere of radius $a$ with charge $Q$. Can you determine the potential energy in this case?

 

[Delta2]

The expression for E2 is also wrong, and one reason is that NOT all the charge Q will flow but some of it until the potential of the two spheres is equal.
If Q' and Q'' are the charges of the sphere and the shell respectively after the process has ended, then we 'll have two equations

1) Conservation of charge will give us the first equation. Before the process the total charge is 2Q. After the process the total charge is Q'+Q''. So it will be (neglecting the small charge that will be stored in the connecting wire)
$Q'+Q''=2Q$ (1)
2) Equality of potentials of the sphere and the shell after the process has ended will give us the 2nd equation
$K\frac{Q'}{a}=K\frac{Q''}{2a}$ (2)

You should solve the system of (1) &(2) to obtain Q' and Q''. Then you must calculate correctly the potential energy BEFORE (when sphere has Q charge and shell has Q charge also) and AFTER (when sphere has Q' charge and shell has Q'' charge). The difference between the two potential energies will be the heat.

 

[haruspex]

NOT all the charge Q will flow

You may wish to review e.g. http://farside.ph.utexas.edu/teaching/302l/lectures/node28.html

 

[Delta2]

You may wish to review e.g. http://farside.ph.utexas.edu/teaching/302l/lectures/node28.html

So you think that reasoning perfectly applies for the specific shape of this setup? Even if it does it isn't clear to me what is the outer surface in this case

 

[haruspex]

So you think that reasoning perfectly applies for the specific shape of this setup? Even if it does it isn't clear to me what is the outer surface in this case

Put the Gaussian surface just inside the outer surface of the shell.

 

[Delta2]

Put the Gaussian surface just inside the outer surface of the shell.

Shouldn't the Gaussian surface "be attached" to conducting material? Because the reasoning goes like "the electric field inside the conductor must be zero (because ...) hence the electric field in the Gaussian surface is zero"

Ok, I think I see now , if the outer shell has a non zero thickness then the reasoning applies with that Gaussian surface, however if it has zero thickness I see a problem.

 

[haruspex]

Shouldn't the Gaussian surface "be attached" to conducting material? Because the reasoning goes like "the electric field inside the conductor must be zero (because ...) hence the electric field in the Gaussian surface is zero"

Ok, I think I see now , if the outer shell has a non zero thickness then the reasoning applies with that Gaussian surface, however if it has zero thickness I see a problem.

Real shells tend to have a nonzero thickness. Besides, if you put the Gaussian surface just inside the inner surface of the shell you still get the result that all charge goes to the shell.