Non conducting shell problem

In summary: This is just the definition of charge density. There are two ways to approach this: find the field at a point and solve for A, or find the integral of the field over the shell and solve for A.In summary, we need to find a value of A such that the field within the charged shell has constant magnitude. There are two ways to do this: find the field at a point and solve for A, or find the integral of the field over the shell and solve for A.
  • #36
rudransh verma said:
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?
rudransh verma said:
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.
 
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  • #37
haruspex said:
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.
 
  • #38
haruspex said:
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.
 
  • #39
rudransh verma said:
it’s a nonconductor and so it’s charge is spread out throughout the shell instead of being just on the outer surface
Yes.
 
  • #40
haruspex said:
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.
 

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  • #41
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.
 
  • #42
haruspex said:
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?
 
  • #43
rudransh verma said:
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).
 
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  • #44
Steve4Physics said:
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.
 
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  • #45
rudransh verma said:
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.)
 

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