Electrostatic energy of the system

1 vote
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Figure shows three concentric conducting spherical shells with inner and outer shells earthed and the middle shell is given a charge q. Find the electrostatic energy of the system stored in the region I and II. I tried as

As the potential of sperical shell I and III are zero , I got two equation
Potential at shell I is zero hence $$K/r[q_1 + (2q/5) + (2q_3 /7)] =0$$
Potential at shell III is zero hence $$2K/7r[q_1 + q + q_3]=0$$

From this I got $q_1 =(-74q/25)$ and $q_3 = (-49q/25)$

edited Feb 8, 2017
I see only one equation here. ... So what do you do next?

Isn't this a trick question? The electric charge on II resides on the outer surface. So no charge is induced on I. There is no electric field between conductors I & II, so no energy is stored in this region.

Do you have a worked solution?
The potential of conductor I should be zero . I assumed charge of conductor I as q$_1$  so there will be charge induced that charge on conductor II would be -q$_1$ on inner side and q+q$_1$ on outer surface.
http://i.imgur.com/YMvIQuG.jpg

1 vote

The arrangement can be viewed as two spherical capacitors in series.

Assume the charges on the inner and outer surfaces of the middle spherical shell are $q_1$ and $q_2$. The total charge on the middle shell is $q_1+q_2=q$. The charges induced on the inner and outer shells are $-q_1$ and $-q_2$ respectively.

The PD between the inner and outer shell of a spherical capacitor is $kQ(\frac{1}{a}-\frac{1}{b})$ where $Q$ is the charge on the inner shell and $a,b$ are the radii of inner and outer shells. So in this case the PDs between inner and middle shells and between middle and outer shells are
$V_1=k(-q_1)(\frac{1}{r}-\frac{1}{2.5r})= -\frac35 \frac{kq_1}{r}$
$V_2=k(+q_2)(\frac{1}{2.5r}-\frac{1}{3.5r})=\frac{4}{35} \frac{kq_2}{r}$.
These PDs are equal (and opposite in sign) because the middle shell is common and the outer shells are at the same potential, so
$-V_1=V_2$
$\frac35 \frac{kq_1}{r}=\frac{4}{35} \frac{kq_2}{r}$
$q_2= \frac{21}{4}q_1$.
Substituting into $q_1+q_2=q$ gives
$q_1=\frac{4}{25}q$
$q_2=\frac{21}{25}q$.

The energies stored in capacitors 1 and 2 are
$U_1=\frac12q_1V_1=\frac{3}{10} \frac{kq_1^2}{r}$
$U_2=\frac12 q_2V_2=\frac{2}{35}\frac{kq_2^2}{r}$

Except for the discrepancy over the sign of $q_1$, which makes no difference to the answers, these expressions agree with the answers you have been given.

answered Feb 8, 2017 by (28,448 points)
edited Feb 8, 2017
What mistake I have done
Sorry, I don't know, because I do not understand how you got your equations. You have not provided enough detail in your explanation.

Why don't you post your solution (in detail) as an answer?  Then I can comment on it.
Now is it fine
No, I still do not understand how you got these equations.
I have just as equated the total potential at shell I and III to be zero .
How did you get the left hand side of each equation?
I want to know what you did. I do not want to spend time looking at other sites to try to figure out what you did. If you are not prepared to explain in detail what you did, I cannot help.
At shell I potential due to charge q$_1$ is Kq$_1$/r , due to q is 2Kq/5r and due to q$_3$ is 2Kq$_3$/7r  . The same way for shell III.
That is the potential at shell III due to charge q$_1$ , q and q$_3$ are 2Kq$_1$/7r , 2Kq/7r and 2Kq$_3$/7r respectively .

Now I have just added these potential and equated to zero
OK I understand what you have done now. You have used the Principle of Superposition.

What you did wrong is that you did not use the correct sign for $q_2$ at conductor III. You assumed charge $+q_1$ is induced on outer surface of conductor I, charges on conductor II are $-q_1$ on inner surface  and $q+q_1=q_2$ on outer surface, and charge $-q_2$ on inner surface of conductor III. However, in your equations you used $+q_2$ as the charge on conductor III when it should be $-q_2$.

After you correct that mistake you get $q_2=\frac{21}{25}q$ and $q_1=-\frac{4}{25}q$.