JR0024 : DCP-16-453-26 : Why did the "centre of charge" method fail while computing the electric potential?

Guru Vishnu

Apr 28, 2020 09:24

Question:

A point charge $q$ is placed inside a conducting spherical shell of inner radius $2R$ and outer radius $3R$ at a distance of $R$ from the centre of the shell. The electric potential at the centre of the shell will be $\frac{1}{4\pi\varepsilon_0}$ times:

(a) $\frac{q}{2R}$
(b) $\frac{4q}{3R}$
(c) $\frac{5q}{6R}$
(d) $\frac{2q}{3R}$

My approach:

The point charge $+q$ induces an equal charge $-q$ on the inner surface of the shell in accordance with the Gauss law. The electric field inside a conductor is zero in electrostatic conditions. So the centre of charge (analogous to centre of mass) of the induced charge on the inner surface of the shell coincides with the position of the point charge $+q$. So, I assumed the negative induced charge to be a point charge of $-q$ placed at the position of the charge $+q$.

Thus potential at the centre of the cavity is solely due to the charge on the outer surface which is nothing but $+q$. So the final answer is $\frac{q}{3R}$. Wow! There's no such option. Can you please tell what went wrong in my method sir?

John Rennie

Let me draw a diagram. Diagrams always help me think :-)

So the potential is:

$$ V = \frac{kq}{R} + \frac{-kq}{2R} + \frac{kq}{3R} $$

Just adding up the potentials for the three charges.

Yes?

Guru Vishnu

Apr 28, 2020 09:41

Ok sir. How do we know the potential due to the charge on the inner surface is $-\frac{kq}{2R}$, the charges are distributed non-uniformly right?

John Rennie

The charges are distributed unevenly, but they are all at a distance $2R$ from the centre.

Guru Vishnu

Fine sir. Since we're concerned with potential, a scalar, this doesn't matter much. I understood this method and this gives the correct answer. But can you please tell why the "centre of charge" method of mine failed?

@JohnRennie

John Rennie

> So the centre of charge (analogous to centre of mass) of the induced charge on the inner surface of the shell coincides with the position of the point charge $+q$.

I'd have to think about whether that's true or not, but in any case it is irrelevant.

Guru Vishnu

It's not obvious to be why it is irrelevant.

John Rennie

You are presumably saying that once we have the centre of charge the potential is just $kQ/x$ where $x$ is the distance to the centre of charge?

Guru Vishnu

Apr 28, 2020 09:56

Yes sir. And here, if $Q$ constitutes of both $+q$ and $-q$, then the potential due to the inner charge and the charge induced on the inner surface is zero (in my method).

John Rennie

Consider a charged spherical shell. No other charges, just a charged sphere of some radius $r$. The centre of charge is the centre of the shell. Yes?

Guru Vishnu

Yes sir.

John Rennie

So if I'm at the centre of the shell the distance from the centre of charge is $x=0$.

Guru Vishnu

Yes sir.

John Rennie

So the potential at that point is $V(0) = kq/0$ ?

Guru Vishnu

Apr 28, 2020 10:00

I agree sir. It must be zero. But here it's becomes an invalid expression. I guess we can tell the same thing for centre of mass in gravitation.

John Rennie

Yes. If you are at the centre of mass that does not mean the gravitational force is infinite.

Guru Vishnu

Ok sir. But we're not computing the potential at the position of charge as the charge is off-centred in this question by an amount of $R$.

John Rennie

Agreed. But my example shows the potential is not just given by $V = kQ/x$ where $x$ is the distance to the centre of charge.

But your method assumes that it is.

Guru Vishnu

That was just based on my experience in Newtonian mechanics. I tried to implement the same here. I don't see how is this much different from the previous cases.

I found an evidence to support my view: I've learnt in Chemistry that polar bonds are the result when the centre of positive charge doesn't coincide with the centre of negative charge. If they coincide there is no dipole moment. This can be related to this question's situation.

@JohnRennie

John Rennie

It's not true in Newtonian gravity either.

Consider a ring of radius $R$, and consider moving in from infinity along the axis of the ring.

The centre of mass lies on the axis, so as we move along the axis the distance to the centre of mass is independent of $R$.

Guru Vishnu

Apr 28, 2020 10:08

Ok sir. Are we considering the gravitational case or the electrostatic case? Or in other words are we concerned about the charge on the ring or its mass?

John Rennie

Gravitational.

Guru Vishnu

Ok so far sir.

John Rennie

(Though it's true for electrostatic as well)

Guru Vishnu

Yes sir. I understand. This is because of their inverse square relationship.

John Rennie

But now take the ring radius to infinity i.e. every part of the ring mass $m$ is infinitely far away.

But since the centre of mass hasn't moved your argument would be that the potential energy as a function of distace along the axis doesn't change.

Guru Vishnu

Apr 28, 2020 10:11

Ok sir. I see. All limitations of cases we consider in Newtonian gravity is equally applicable to electrostatics we're considering now.

John Rennie

It's just not true that the potential is a function only of the distance from the centre of mass/charge.

Guru Vishnu

Ok sir. What other factors are involved here?

John Rennie

The potential is proportional to $\int dm/r$ or $\int dq/r$

You need to evaluate that integral to calculate it.

Guru Vishnu

Ok sir. I'll take an alternate approach to this.

@JohnRennie: Do you agree that, when we place a charge +q and -q very close to each other, the effects of one charge cancels the effect of the other? Or in other words, the field due to the combination and hence the potential is zero? In short, the combination acts like a neutral mass. Ok sir?

John Rennie

Only when we are far away compared to the spacing between the charges

Guru Vishnu

Apr 28, 2020 10:23

Yes sir. Here let's consider somehow they're exactly in the same position. Ok sir?

John Rennie

What we get is a dipole field where the dipole is equal to $p = qd$, where $d$ is the distance between the dipoles.

If they're exactly the same position then $d = 0$ so $p = 0$ and the dipole field is everywhere zero. Yes.

Guru Vishnu

Fine sir.

Do you agree that a ring uniformly charged with $+q$ acts as a point charge placed at the centre if the dimensions of the ring are small?

John Rennie

@GuruVishnu the ring acts as a point charge only if the distance to the ring is large compared to the radius of the ring.

Guru Vishnu

Ok sir. The same is applicable for "centre of mass" and gravitation problems. Let's forget this and move on to the next point. Ok sir?

John Rennie

I'm not sure what we are arguing about. There are certainly cases where it is a good approximation to treat a charge distribution as a single point charge located at the centre of charge.

But likewise there are cases where it is not a good approximation.

Guru Vishnu

Apr 28, 2020 10:29

I agree sir. I brought up this example to relate with the classic centre of mass. A bit of deviation from the main point which I causing confusion to me:

Things we know:

- charge on the outer surface is uniform

- charge on the inner surface is non-uniform

John Rennie

> charge on the outer surface is uniform

No.

The charge on the outer surface will not be uniform.

Guru Vishnu

So is the following incorrect?

I have some reasons (right or wrong) to support why the charge on the outer surface is uniform.

John Rennie

Hmm, would I swear under oath that the charge on the outer surface is non-uniform?

I certainly can't see any reason why it should be uniform.

Guru Vishnu

Ok sir. Then it's my turn to prove this :-)

BTW the image is from Halliday, Resnick, Walker borrowed from an answer on the main site.

The red dotted surface is the Gaussian surface and in electrostatics the field is zero and hence the flux. The field is zero implies, the outer surface or its charge will never know whether a charge is inside or not in the cavity. It doesn't even know whether any cavity exists.

And hence I'm 100% sure the charge on the outer surface is uniform.

Can you support your views @JohnRennie sir?

5

A: Electric field inside a conductor and induced charges

The point is the charges on the outside reorganize themselves so the net field is $0$ inside the conductor. The charge distributions on the inside and outside surfaces need not be constant and in general will be quite messy unless the geometry is simple. In the example below of a source charg...

> In the example below of a source charge off centre inside a hollow sphere, notice how the positive charges on the inside surface are not uniformly distributed, but how they are uniformly distributed on the outside surface.

John Rennie

Apr 28, 2020 10:47

Hmm, the answer just states the charge is uniform without giving a proof.

Guru Vishnu

@JohnRennie But what about my explanation?

John Rennie

I'm not saying you're wrong, because I'm uncertain of this, but you seem to be saying the field at the red line is zero therefore the outer charge distribution is even. But you could also argue that the field at the red line is zero only because the charge at the outer surface is uneven and balances out the uneven inner charges.

Guru Vishnu

Apr 28, 2020 11:58

@JohnRennie: Hi sir. When you're back here, please see the following post (which I wrote a long time ago):

1

A: Why is the charge distribution on the outer surface of a hollow conducting sphere uniform and independent of the charge placed inside it?

When you place a negative charge $-q$ inside a hollow conducting sphere, $+q$ amount of charge is induced on the inner surface of the hollow sphere. This is because of the fact, in electrostatics the net electric field inside a conductor must be zero. If we consider a Gaussian surface as shown in...

My answer is a basic (intuitive) one based on my understanding.

A rigorous answer, which I don't understand anything is given by ZeroTheHero:

1

A: Why is the charge distribution on the outer surface of a hollow conducting sphere uniform and independent of the charge placed inside it?

This is best tackled by first considering potentials. Because the sphere is a conductor, the outer surface of the sphere is an equipotential, and of course outside the sphere one ought to solve $\nabla^2 V=0$ in spherical coordinates to obtain the potential everywhere, from which we would deduce...

John Rennie

Apr 28, 2020 13:24

What do you want to discuss?

Guru Vishnu

It's regarding our previous discussion sir. Did you read my answer?
I hope that explains why the charge on the outer surface is uniform.

John Rennie

I haven't had time to read through ZeroTheHero's answer, but if he says the charge on the outer shell is uniform I am prepared to believe him.

Guru Vishnu

Ok sir. Fine. Next shall we move on to analyse why considering the "centre of charge" method fails?

My doubt is why is the potential dependent both on the internal charge and the internal induced charge in addition to the external induced charge as opposed to only the external induced charge.

John Rennie

Short of actually doing the calculation, which would be hard, I don't know how to convince you. Given that we can easily come up with systems where the centre of charge method fails I don't understand why you think it might work in this case.

Guru Vishnu

Apr 28, 2020 13:43

Ok sir. No problem. The reason why I think "centre of charge" method will work here is just based on how I'm used to "centre of [quantity]" in different contexts. I don't know why it doesn't work for potentials like this case, and I don't know why it shouldn't work as I don't see potentials to be much different than other quantities.

John Rennie

But there loads of other simpler cases where it doesn't work.

Guru Vishnu

Apr 29, 2020 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

Apr 29, 2020 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

Apr 29, 2020 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 :-)

JR0024 : DCP-16-453-26 : Why did the "centre of charge" method fail while computing the electric potential?

♦ Particle Accelerator

Participants