Particle Accelerator

Guru Vishnu

04:24

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

Are you free now sir?

John Rennie

@GuruVishnu yes

Guru Vishnu

I found a simple example where the centre of charge fails - when we place a charge +q at the centre of hollow spherical shell, -q amount of charge is uniformly induced on the inner surface and +q amount of charge is uniformly induced on the outer surface.

Of course the centre of charge of the inner -q coincides with the point charge +q. But I think we can use the centre of charge analysis only beyond the boundary of both charge distributions or in other words, it can be used only at points outside the inner sphere.

The electric field and the potential inside the cavity is not zero and constant respectively as suggested by the centre of charge analysis. However, the field is due to the point charge alone and potential is due to both the variable potential due to point charge and constant potential due to the -q amount of charge on the inner surface and +q amount of charge on the outer surface.

John Rennie

Yes

The centre of charge fails outside as well, though in fact it leads you to multipole analysis which is an interesting subject.

Guru Vishnu

04:36

@JohnRennie So the main point I was missing was - using the centre of charge analysis within the boundary. Or in other words when we apply this method, on converting -q amount of inner induced charge to a point charge of same value at the centre at an internal point, the charge distribution passes through the point under our attention.

John Rennie

Yes

Guru Vishnu

@JohnRennie Wow. Really sir? I thought it would work for outside points. But again I'm incorrect :-)

Here it does seem to work, in both simple cases - point charge at the centre and point charge not at the centre.

John Rennie

@GuruVishnu consider a positive and negative charge separated by a distance $d$. This creates a dipole $p = qd$.

If you use the centre of charge approach the field should be zero because the two charges would cancel each other out.

But of course the field is a dipole field not zero.

Guru Vishnu

@JohnRennie Are you using the same type of formula that we use to find centre of mass sir? I think here the centre of positive and negative charges don't coincide. If they coincide do you refer to the origin as the centre of the charge distribution, sir?

John Rennie

If you take any arbitrary distribution of charges it creates a field that can be written as a sum of a monopole, dipole, quadrupole, octopole, etc.

Your centre of charge idea is exactly the monopole component, and in many cases this is the dominant component so the centre of charge is an excellent approximation.

Where it fails is when the other components become large. A pure dipole is an extreme example of this since the monopole component is zero.

Guru Vishnu

04:42

@JohnRennie Ok sir. Sounds like expressing $\sin x$ (or other functions) as Taylor expansions. So more the number of terms like octupole, 14-pole, 140000-pole, … the better the approximation. Am I right, sir?

John Rennie

@GuruVishnu yes

@GuruVishnu the position of the centre of charge is the same as the centre of mass i.e.

$$ \mathbf R = \frac{1}{Q} \int q(\mathbf r) d\mathbf r $$

Guru Vishnu

Fine sir. I see how it's zero for a dipole. I think by "centre of charge" analysis I meant something different - analysing whether the centre of all positive charges is same as all centre of negative charges where I compute the integral only for charges of one type.

@JohnRennie: Is there anything else I need to know regarding this for now, sir?

John Rennie

No, multipole analysis isn't needed for the JEE.

Guru Vishnu

Ok sir. Thank you :-)

(removed)

Guru Vishnu

05:41

@JohnRennie: Hi sir. Can you please ping me after Aladdin's doubt?

John Rennie

@GuruVishnu I think we are finished.

Guru Vishnu

Ok sir. I'm having a doubt in the following question:

Question:

Capacity of a spherical capacitor is $C_1$ when inner sphere is charged and outer sphere is earthed and $C_2$ when inner sphere is earthed and outer sphere is charged. Then $C_1/C_2$ is ($a=$radius of inner sphere, $b=$radius of outer sphere)

(a) $1$
(b) $a/b$
(c) $b/a$
(d) $\frac{a+b}{a-b}$

My doubt:

First of all how can capacitance depend on which of the two plates is connected to the earth? It's a property of the capacitor as a whole right? And must be $\frac{4\pi\varepsilon_0 ab}{b-a}$ right sir?

(End of message)

John Rennie

That puzzles me as well ...

Guru Vishnu

Fine. I forgot to tell. I initially chose option (a) $1$, but the key and the solution states it's $a/b$ (option b). Do you wish to see the solution provided by the book sir?

John Rennie

Yes

Guru Vishnu

05:46

They say that:

$$C_1=4\pi\varepsilon_0\frac{ab}{b-a}$$

and,

$$C_2=4\pi\varepsilon_0\frac{b^2}{b-a}$$

Then they divide this to give $a/b$. But I don't see how as I mentioned earlier.

@JohnRennie: And there was no explanation for the formula for $C_2$. I understood the derivation of $C_1$ which is the normal capacitance of a spherical capacitor which I learnt a long time ago but the second formula seems different. Why do they take the radius of the inner sphere to be $b$ in the numerator and to be $a$ in the denominator?

John Rennie

Don't know. I'll have to think about it.

Guru Vishnu

Ok sir.

John Rennie

06:01

If you earth one of the plates you fix the potential of that plate to be zero. e.g. if you earth the outer plate the potential has to be zero meaning the field outside the outer plate must be everywhere zero.

Guru Vishnu

Ok sir. I understand earthing means the potential of the conductor must be zero (same as Earth) but why must the field be zero outside?

John Rennie

If you earth the inner plate then the potential of the inner plate must be zero, so if you integrate radially outwards from the inner plate to infinity the result must be zero. That means the field outside the outer plate is no longer zero i.e. there must be a net non-zero charge on the two spheres.

@GuruVishnu if you earth the outer plate then its potential is zero, so if you integrate the field radially from the outer plate out to infinity the result must be zero. But there is no charge outside the outer plate, so the only way for the integral to be zero is if the field outside the outer plate is everywhere zero.

Guru Vishnu

Ok sir.

John Rennie

So which sphere you earth does change the field, and therefore presumably changes the potential difference between the two spheres

i.e. for a given charge the voltage will vary and hence C = Q/V will change.

Guru Vishnu

Ok sir. So are we going to fix the potential difference between the two plates and find the ration of charges in the capacitor system to find the required ratio of capacitances. This seems simple.

John Rennie

06:10

Yes

Or keep the charge constant and find the ratio of the voltages.

Guru Vishnu

Ok sir. Thank you. Let me try that :-)

@JohnRennie: Again it comes to be $1$ sir. For both the cases, I get $$V=kq/b - kq/a$$ where $q$ is the charge on the inner surface of the outer sphere and outer surface of the inner sphere.

John Rennie

With the inner sphere earthed the charges won't be the same, or at least I don't think so.

Guru Vishnu

Fine. That was just based on Gauss law.

So charges on the facing sides is equal in magnitude and opposite in sign.

John Rennie

Suppose we gave a charge $q_b$ on the outer sphere and $q_a$ on the inner sphere.

Then the potential at the outer sphere is:

Guru Vishnu

$$V(b)=kq_b/b+kq_a/b$$

John Rennie

06:22

And the potential at the inner sphere is:

$$ V(a) = kq_b/b+kq_a/a $$

And by earthing the inner sphere we impose the condition that $V(a) = 0$

Guru Vishnu

Ok sir.

John Rennie

So we get:

$$ q_a = -\frac{a}{b} q_b $$

$$ V(b) - V(a) = \frac{kq_a}{b} - \frac{kq_a}{a} $$

Guru Vishnu

Ok sir. But shouldn't we just be considering only the charges on the facing sides of the spherical capacitor?

John Rennie

Let's go with this and see what happens ...

Guru Vishnu

Ok sir.

John Rennie

06:28

$$ V(b) - V(a) = \frac{kq_a}{b} - \frac{kq_a}{a} = kq_a \frac{a-b}{ab} $$

or substituting for $q_a$ we get:

$$ V(b) - V(a) = kq_b \frac{a}{b} \frac{a-b}{ab} $$

$$ V = kq_b \frac{a-b}{b^2} $$

$$ C = \frac{Q}{V} = \frac{b^2}{k(a-b)} $$

Guru Vishnu

Ok sir. It's exactly the same equation provided in the solution. Now I get the mathematical part of this. But the result is quite surprising.

John Rennie

Taking the ratio of this with the value for the outer sphere earthed is going to give us the correct answer, isn't it?

Guru Vishnu

Yes sir.

Does it matter whether we connect the inner or outer surface to the earth for both inner and outer sphere sir? I don't think so.

Both would have the same effect as potential is uniform throughout in a conductor.

John Rennie

If we connect the outer sphere to earth we know the potential at the outer sphere must be zero so:

$$ V(b) = \frac{kq_b}{b} + \frac{kq_a}{b} = 0 $$

So $q_b = -q_a$

That's the difference.

Guru Vishnu

Yes sir. I understand that it will give the regular spherical capacitance formula.

My doubt was regarding the following formula which I read before asking you:

8

A: Capacitance of spherical capacitor when earthed

Consider the following cases in relation to your question: Inner sphere is grounded. a) grounding the outer surface of the inner sphere If you ground the outer surface of the inner sphere, the inner sphere becomes irrelevant and you get single spherical capacitor (the other one at infinity) ...

John Rennie

06:36

With the outer sphere earthed the charges are equal and opposite. With the inner sphere earthed some charge flows from earth onto the inner spere and the charges become unequal.

Guru Vishnu

It seems there is a difference between connecting the inner and outer surface of the inner sphere to earth. But I don't see how, sir.

John Rennie

There is a danger in taking too much time to try and get an intuitive understanding. At some point you have to give up and move on.

Guru Vishnu

I agree sir. I'm not worried about an intuitive understanding but that answer seems to suggest connecting the two surfaces of a shell has different effects on the capacitance.

When you find time, can you please see the answer linked above?

Guru Vishnu

07:09

@JohnRennie: Hi sir.

John Rennie

@GuruVishnu hi

Guru Vishnu

Can you please see the following? :

33 mins ago, by Guru Vishnu

8

A: Capacitance of spherical capacitor when earthed

Consider the following cases in relation to your question: Inner sphere is grounded. a) grounding the outer surface of the inner sphere If you ground the outer surface of the inner sphere, the inner sphere becomes irrelevant and you get single spherical capacitor (the other one at infinity) ...

It seems the two sub cases under the main two cases is irrelevant but the reason seems to be valid.

And that user manages to get the final expression we got.

John Rennie

His answers are correct but he doesn't give any working and that makes his answer rather unsatisfactory to me.

Guru Vishnu

Yes sir. But doesn't that seem different - connecting either the inner surface of the outer plate or the outer surface of the inner plate must have the same effect right?

John Rennie

Yes. The plate is a conductor, so it doesn't matter what point on the plate is connected to earth.

Guru Vishnu

07:19

Fine. Thank you for the clarification.

Then how did the answerer manage to bring two extra different results?

John Rennie

His only different result comes when you ground both plates

In that case the whole system just behaves like a single sphere so the capacitance is the self capacitance of a sphere with radius $b$.

Guru Vishnu

@JohnRennie Grounding both the plates? Not that sir. I meant he managed to get two different results each for grounding the outer surface and the inner surface of the inner sphere.

John Rennie

Oh, I see, his case 1a ? I don't know what he means there.

Guru Vishnu

Ok sir. We're on the same track. No issues!

Thank you.

John Rennie

Not everything that is upvoted on the SE is correct. That applies to some of my answers as well :-)

Guru Vishnu

07:28

:-)

Guru Vishnu

08:31

@JohnRennie: Hi sir. Are you free now?

John Rennie

@GuruVishnu yes

Guru Vishnu

I'm having a doubt in one of your steps regarding our previous discussion:

2 hours ago, by John Rennie

or substituting for $q_a$ we get:

Here, why did we substitute $q_b$ in place of $q_a$?

If we had proceeded with:

2 hours ago, by John Rennie

$$ V(b) - V(a) = \frac{kq_a}{b} - \frac{kq_a}{a} = kq_a \frac{a-b}{ab} $$

Then the final result would have been $1$ as I said before which is of course incorrect.

John Rennie

We are working out the capacitance by placing a charge $q$ on the capacitor and finding the corresponding voltage, then using $C = Q/V$.

Guru Vishnu

Yes sir. I understood that point.

John Rennie

So the question is what should we use for the value of $q$? Should it be the charge on the inner or outer sphere?

Since the charges are different when the inner sphere is earthed, we will get a different answer depending on what charge we use.

Guru Vishnu

08:35

@JohnRennie I think, it must be the charge on the facing sides, or generally, the charge on the outer surface of the inner sphere.

2

Q: How is the "charge on a capacitor" defined when two plates are unequally charged?

In Concepts of Physics by Dr.. H.C.Verma, in the chapter on "Capacitors", in page 144, under the topic "Capacitor and Capacitance" the following statement is given: A combination of two conductors placed close to each other is called a capacitor. One of the conductors is given a positive char...

That was based on the answers to the question above.

Or what I mean is, here we're interested in the charge on the facing surface more than the overall charge...

John Rennie

The way I would look at this is: suppose we start with neither sphere earthed and add charges $q$ and $-q$ in the usual way. Then we earth one of the plates, find the new voltage and calculate $C = q/V$ using the charge we originally put on the spheres i.e. ignoring any change due to charge flowing to or from earth.

Guru Vishnu

Ok sir. Then the value of $C$ would be independent of which of the two plates is earthed. Right?

John Rennie

No.

Guru Vishnu

But this is what I can interpret from "ignoring any change due to charge flowing to or from earth"; Can you please tell which point I'm misinterpreting, sir?

John Rennie

Suppose we put a charge $q$ on the capacitor then earth the outer sphere. No charge flows to or from earth and the potential difference between the spheres remains unchanged.

OK so far?

Guru Vishnu

08:39

Ok sir.

John Rennie

Now we put the same charge on the capacitor then earth the inner sphere. This time the charge on the inner sphere changes because some charge flows between the inner sphere and earth. This changes the potential difference $V$ between the spheres, so it changes the value of $q/V$. Yes?

Guru Vishnu

Yes sir.

John Rennie

So the ratio $q/V$ is different depending on which of the spheres we earth, so the capacitance is different.

Guru Vishnu

What if $q$ and $V$ both change in a way $C$ remains constant, sir?

John Rennie

$q$ is the charge we applied before we connected either sphere to earth. It is a constant.

That's how we are defining $q$. It is the original charge, not the charge after some current has flowed to or from earth.

Guru Vishnu

08:46

Fine sir. I think that's where I'm facing confusion. Based on the question linked above, Andrew states charge on a capacitor is the amount of charge that flows when the two plates are connected. This means the charge must be the one on the facing surfaces. I don't understand why we need to consider $q$ as the one present at the beginning.

John Rennie

Oh wait ...

Guru Vishnu

Ok sir.

John Rennie

Suppose we place a charge $q$ on the capacitor then earth the inner sphere. The charge on the outer sphere remains at $q_b=q$ while the charge on the inner sphere changes to $q_a = -\frac{a}{b}q_b$. Yes?

Guru Vishnu

Yes sir.

John Rennie

Now suppose we connect the plates. When we connect the plates both plates are earthed so both plates have a charge of zero. Yes?

Guru Vishnu

08:55

Yes sir.

John Rennie

So the charge that flows between the plates when they are connected must be $q_b$, because the outer sphere started with a charge $q_b$ and ended with a charge $0$ and it is only connected to the inner sphere so that charge $q_b$ must have flowed between the spheres.

And $q_b = q$ i.e. the charge originally applied.

Guru Vishnu

Ok sir. I agree. Thank you very much :-)

John Rennie

So whether you define the $Q$ in $Q/V$ as the charge originally applied or the charge that flows between the spheres you get the same answer.

Guru Vishnu

Fine. That works here. Does it mean it would work anywhere?

John Rennie

I don't know. To be honest it had never occurred to me before.

Guru Vishnu

09:00

Ok sir. No problem. Let me go through our conversation once again from the beginning and attempt this problem. Thank you.

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♦ Particle Accelerator