How Do We Calculate Surface Charge Density on a Non-Conducting Shell?

[Steve4Physics]

Why is the question telling us that field inside to be uniform. Is there a field inside the shell. How do we calculate that?

If I may chip in...

The question is asking “What value of ‘A’ makes the field (between r=a and r=b) ‘uniform’?

Note, the term ‘uniform’ is wrong. For a field to be uniform, both its magnitude and direction must be the same everywhere in the field. Pictorially we represent this by parallel, evenly spaced, field lines. We don’t have a uniform field in this question.

The question should say ‘What value of A makes the magnitude of the field (between r=a and r=b) uniform?” This has already been noted by @haruspex in Post #2.
_____________

Yes there is a field inside the shell (between r=a and r=b). The formula for it is the one you give in your image in Post #14. The formula becomes very simple after you substitute $A= \frac {Q}{2\pi a^2}$
____________

A few points about'A' that seem to be a source of confusion...

‘A’ may, or may not, have a physical meaning. Even if it does, you don’t need to know it to answer the question.

‘A’ is simply a number you are being asked to find. It is the value which makes in the magnitude of the field (everywhere between r=a and r=b) the same. That's because in the equation for E, for the correct value of A, the term containing 'r' disappears.

I imagine that if A is too big, the field going from from r=a to r=b increases. And if A is too small, the field going from r=a to r=b decreases.

A = charge-density x r so is is easy to see that A has units of C/m². A is clearly is not an area.

 

[haruspex]

Why is the question telling us that field inside to be uniform.

I explained in post #2 that the question means the field has constant magnitude within the walls (a<r<b) of the shell, and in post #17 that the direction of the field is radial, so not uniform.

Is there a field inside the shell.

When discussing shells one has to be careful to avoid ambiguity. By "inside the shell" here do you mean r<a or a<r<b?
Either way, yes there is a field. For r<a, the field is that generated by the isolated charge at the centre.
For a<r<b, you calculated the field.

 

[haruspex]

‘A’ is simply a number you are being asked to find.

Well, it has dimension, so I would not say it is just a number.

 

[Steve4Physics]

Well, it has dimension, so I would not say it is just a number.

Good point. In retrospect, 'value' would have been a better choice of word.

 

[rudransh verma]

I explained in post #2 that the question means the field has constant magnitude within the walls (a<r<b) of the shell, and in post #17 that the direction of the field is radial, so not uniform.

I have a confusion. We have calculated firstly the amount of charge from a to r. The field at r (at surface of a shell from a to r)due to this charge is like a charge at the center. And we have used the eqn E=1/4pie0q/r^2. I mean how else will we use this eqn if the field is not at the surface of shell from a to r. I guess we can say this is a smaller shell. We can say then that E is inside the shell from a to b. It’s a non conductor so charges don’t move to the surface of shell.

Either way, yes there is a field. For r<a, the field is that generated by the isolated charge at the centre.
For a<r<b, you calculated the field.

Then we calculated the field due to both charges Q and q at r which will simply add up vectorially and this will be the net field.
And of course field will be uniform in magnitude because of symmetry.

Right ?

 

[haruspex]

how else will we use this eqn if the field is not at the surface of shell from a to r.

Sorry, I do not understand the question. We wanted to find the field at radius r, which is the same as saying the field at the surface of the shell radius r.
What else have you in mind?

field will be uniform in magnitude because of symmetry.

It is uniform in magnitude with respect to radius because of the value chosen for A, and everywhere radial in direction because of the symmetry.

 

[rudransh verma]

We wanted to find the field at radius r, which is the same as saying the field at the surface of the shell radius r.

Yes that is what I wanted to say. We can imagine there is a imaginary shell of outer radius r and we have to find the field at its surface.

 

[rudransh verma]

What else have you in mind?

I want to sort out one thing. Can we find field of shell from r to b in place of a to r(a to b I think will yield nothing) and thus get the right answer?

And I have been mentioning a couple of times it’s a nonconductor and so it’s charge is spread out throughout the shell instead of being just on the outer surface. You haven’t pointed out anything so I assume I am right.

 

[haruspex]

it’s a nonconductor and so it’s charge is spread out throughout the shell instead of being just on the outer surface

Yes.

 

[rudransh verma]

Yes.

Shell theorem.
This is a spherical shell which I assume is very thin. That is why we took R as radius. Now we are trying to find E due to this shell at S2. I have done it attached in picture. Here we have taken out E as a constant. That means E produced at S2 is constant at all points.
I want to ask how?
I know field will be constant if it’s a point charge at the center but how is field constant due to a charged shell.

Also I have tried to do the above problem another way but it’s not coming. A is coming zero is wrong. How can we do it by first finding dE and then summing up all dEs.

 

[Steve4Physics]

Hope I’m not being intrusive. I think this might help, but apologies if not...

@https://www.physicsforums.com/members/rudransh-verma, here’s a different problem. If you think about it, maybe it will help you deal with the original problem.

You have a sphere, radius R, containing charge Q, spherically symmetrically distributed. (It doesn’t matter if Q is spread throughout the sphere or just on its surface.)

We know that at the surface of the sphere, the field’s magnitude is $\frac {kQ}{R^2}$, where $k = \frac {1}{4\pi \epsilon_0}$.

Outside the sphere, a distance r from the sphere’s centre, the field’s magnitude is $E = \frac {kQ}{r^2}$. So (for r≥R), E decreases as r increases, in accordance with the inverse square law.

Problem: You are allowed to surround the sphere with any amount of additional charge, distributed in any way you want. Is there a way to do this so that when r increases, the magnitude of the field outside the sphere stays the same? This would make the magnitude of the field outside the sphere always equal to $\frac {kQ}{R^2}$, no matter what the value of r is.

 

[rudransh verma]

Anything else?

I don’t understand why we take a and r as limits when we are given a and b in question when finding A. Can I not achieve the same answer by taking a and b?

 

[Steve4Physics]

I don’t understand why we take a and r as limits when we are given a and b in question when finding A. Can I not achieve
he same answer by taking a and b?

The strategy we use is:
i) find a formula for |E| as a function of r (where a≤r≤b);
ii) using a bit of algebra, find an expression for 'A' which removes r from the formula for |E|.

We find a formula for |E| as a function of r in order that we can then figure out how to choose A so our formula no longer contains r (which makes |E| is independent of r).

I can’t see any other sensible strategy.

It's worth noting that the constant value of |E| will be whatever |E| is when r=a. So we know (e.g. using Gauss’s theorem) that $|E| =\frac {q_0}{4 \pi \epsilon_0 a^2}$ where $q_0$ is the central charge.

It's also worth noring that the value of b is irrelevant! The answer is true for any value of b (as long as b>a).

 

[rudransh verma]

The answer is true for any value of b (as long as b>a).

You mean there is no need for b anywhere in equations.

 

[Steve4Physics]

You mean there is no need for b anywhere in equations.

Yes - no need for b!

(If there were an additional part of the question such as 'What is the total charge?", then you would need value for b.)