# Homework Help: Heat energy from a charged configuration

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1. Apr 5, 2016

### alivedude

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

I have two spherical metallic shells with radius $3a$ and $a$, the little one is placed inside the larger so that the center of the little one is at a distance of $a$ from the center of the larger. The outer one has a charge $3Q$ and the one inside has a charge of $Q$ (both are positive).

Now if we are grounding the outher shell with a thread that has a resistance R some energy will be released as heat, calculate the heat!

2. Relevant equations

This is the problem

3. The attempt at a solution

The problem is 15 years old and in the solution they are referring to some litterature that we dont use today so I got the answer itself but no solution to follow.

This is my thoughts so far:

When we set this configuration up we are storing some energy in the electric field. This energi can be calculated as

$W = \frac{\epsilon_0}{2} \int_{ {\rm I\!R}^3} ( E_1^2 + E_2^2 + 2\textbf{E}_1 \cdot \textbf{E}_2 ) dV$

and when we are grounding the outher shell, all of these should be released somehow, isn't it all heat then? Is the $R$ only there to confuse me?

2. Apr 5, 2016

### TSny

You have the right idea about integrating the square of the field to get the stored electric energy. And, yes, |ΔW| will equal the heat.

Can you sketch (qualitatively) the electric field lines everywhere for the case before the outer sphere is grounded and for the case after the outer sphere is grounded? This will help you see how to set up the integral for |ΔW|.

3. Apr 5, 2016

### alivedude

I can't make any sketch at this moment, I don't have any program on this device. But I calculated $W$ and got $W= \frac{2 Q^2}{3 \pi \epsilon_0 a}$ and this agrees with the answer. So if I haven't done any mistakes I guess that the take-away from this problem is that all of the energy stored in the field will equal the heat and its independent of the resistance, correct?

4. Apr 5, 2016

### TSny

Not all of the energy stored in the field is converted to heat. After the outer shell is grounded, there will still be some electric field energy left in the system. The heat produced is the difference Winitial - Wfinal.

5. Apr 5, 2016

### alivedude

Ok I gave up pretty quickly and tried to do it all in another way and while I was doing it I realised something. Is the formula I stated above really true for conducting shells, like we have in these problems? Because the field is zero inside the conductors and if we are standing outside the whole thing we would have no clue about where the electric charge inside is located, we would only knew that the outer shell is charged with $4Q$. It might be true even for this case but I couldn't get it right, would be really happy if someone could sort it out.

Anyway I solved the problem like this instead:

Gauss told me that the electric field outside is
$E(r)=\frac{Q}{\pi \epsilon_0 r^2}$
and for the potential we have

$V(3a) = V(\infty)+\int_{ra}^{\infty} \textbf{E} \cdot d\textbf{l} = \frac{Q}{3\pi \epsilon_0 a}$
Now if we ground the outer sphere the potential on the inner sphere will decrease by $V(3a)$ and we can calculate the heat like

$W_{Heat} = W_i-W_f = \frac{1}{2}3Q\frac{Q}{3\pi\epsilon_0a}+\frac{1}{2}V_{inside}Q-\left[0+\frac{1}{2}(V_{inside}-\frac{Q}{3\pi\epsilon_0a})Q \right] = \frac{2}{3}\frac{Q^2}{\pi \epsilon_0 a}$

And this should be correct, at least it is the same as the answer in my papers.

6. Apr 5, 2016

### TSny

OK, that looks good. Note that your solution is based on knowing that the field outside the outer (ungrounded) shell is spherically symmetric so that you can use Gauss' law to find E outside the outer shell. Also, you have used the fact that grounding the outer shell does not change the electric field between the two shells.

So, the electric field energy $\frac{\epsilon_0}{2}\int{E^2 dV}$ between the two shells does not change when grounding the outer shell. Thus the loss of electric energy is due just to the loss of field outside the outer shell. You might see if you get the answer by doing the integral $\frac{\epsilon_0}{2}\int{E^2 dV}$ for the initial field outside the outer shell.

7. Apr 11, 2016

### alivedude

Oh yes, thats right! At least its solved now, thank you for your help :)