# Heat energy dissipated between two conducting shells

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1. Aug 10, 2015

### Arka420

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

A conducting sphere of radius a is surrounded by a neutral conducting shell of radius b (b > a).Connections are provided as shown in diagram. Initially, the sphere has a charge Q. The switch S1 is opened and then closed. The switch S2 is then opened and closed. Finally, the switch S3 is closed. What is the heat energy dissipated in the system after S3 is closed?

2. Relevant equations
E = k Qq/r2
V = k Qq/r

3. The attempt at a solution
Work done is equal to the product of the charge and the potential difference. So it's my guess that, after switch S3 is opened, the product of the charge difference and the potential difference between the sphere and the shell IS the heat energy produced. However,this idea has only confused me all the more,and I don't even have a glint of how to proceed with the sum. (I am familiar with Gauss's law for electrostatics).

2. Aug 10, 2015

### SammyS

Staff Emeritus
I think you have a mistake in terminology.

When a switch is closed, current may flow through it . When it's open, no current can flow through it.

By the way: Your relevant equations are in error. They should not have Q⋅q, only Q or q. Not both.

Last edited: Aug 10, 2015
3. Aug 10, 2015

### rude man

Does current flow when S1 is closed?
Does current flow when S2 is closed?
Does current flow when S3 is closed?
What is q in your formulas for E and V?

4. Aug 18, 2015

### Vibhor

I think no heat energy is dissipated after switch S3 is closed .Closing switch S3 doesn't make any difference to the previous state.

But $\frac{KQ^2}{2b}$ Joules of heat energy is liberated after switch S1 is closed and $\frac{KQ^2}{2}(\frac{1}{a}-\frac{1}{b})$ Joules after switch S2 is closed (entire charge on the shells flows down to earth) .

Do you agree with the above results @rude man ?

Last edited: Aug 18, 2015
5. Aug 18, 2015

### rude man

agreed.
how come?

6. Aug 18, 2015

### Vibhor

Electric potential energy of the system when all switches were open = $\frac{KQ^2}{2a}$

Electric potential energy of the system when switch S1 is closed = $\frac{KQ^2}{2}(\frac{1}{a}-\frac{1}{b})$

The difference of the two is the heat energy dissipated .

7. Aug 18, 2015

### rude man

In order for there to be current flowing thru S1, don't you have to have charge on the shell before it gets grounded?
And, why is the potential energy of the system changed when S1 is closed?

8. Aug 18, 2015

### Vibhor

According to you @rude man what should be the value of heat dissipated when switch S1 is closed ?

Last edited: Aug 18, 2015
9. Aug 18, 2015

### rude man

I think you should reason that out for yourself after I gave you my last response.

10. Aug 18, 2015

### Vibhor

If you are suggesting that since there is no charge on the shell , no charge flows when switch S1 is closed , then I beg to differ .

Last edited: Aug 18, 2015
11. Aug 18, 2015

### rude man

I disapprove of what you say, but I will defend to the death your right to say it.
r m

12. Aug 18, 2015

### TSny

I'm a bit confused with the wording of the problem, as was SammyS.

The figure shows all three switches initially in the "open" position. So, it is odd to say "S1 is opened and then closed" when S1 was already in the open position.

From later posts, it seems that the following might be the intended interpretation. All switches are initially open with Q on the inner sphere. S1 is then closed and remains closed. S2 is then closed and remains closed. Finally, S3 is closed. Is this correct?

Last edited: Aug 18, 2015
13. Aug 18, 2015

### Vibhor

Yes.

14. Aug 18, 2015

### TSny

15. Aug 18, 2015

### Vibhor

Great !!!

Thanks a lot

16. Aug 18, 2015

### rude man

@T - the shell has no charge on it before S1 is closed, and has no charge on it after S1 is closed, so how can there be current thru S1? Where is the closed circuit path for that current?
I don't see any change in the E field anywhere after S1 is closed. The sphere is still charged, the shell's inside still has induced & opposite charges on it, and the shell has zero net charge on it always.
r m

17. Aug 18, 2015

### TSny

The net charge on the outer shell is zero, but there will be induced charge on the inner surface and opposite induced charge on the outer surface of the shell. The charge on the outer surface "runs to ground" when S1 is closed.

There doesn't need to be a closed circuit path in order for charge to move between two conductors of different potential. (Think of the ground as a very large conductor at zero potential while the outer sphere is initially at nonzero potential.)

The field outside the larger sphere disappears when S1 is closed.

18. Aug 18, 2015

### rude man

Thanks T. Must mull this over.
r m

19. Aug 18, 2015

### rude man

@Vibhor: TSny is one of our top advisers here at pf and you're right to go with him. I need to convince myself but I already know he's right! Apologies for misleading you.
r m

I suppose the whole thing could have been approached by energy conservation:
How much energy to charge the sphere (all switches still open:)
answer: kQ/a2. (The shell does not alter the E field from infinity to the sphere.)
How much energy left after all switches are closed:
answer: zero, since the sphere and shell are both at zero potential.
So total dissipated energy is just kQ2/2a, agreeing with T and the OP.
Further comment welcome.

Last edited: Aug 18, 2015
20. Aug 18, 2015