Please explain this suggested solution for the field inside a charged hollow sphere

  • #1
tellmesomething
329
38
Homework Statement
Theres a hollow sphere whose ## \sigma ## = ## \sigma_0 cos \theta ## where ##\theta ## is the polar angle/THE angle made with the x axis. Find field inside shell.
Relevant Equations
Expression of electric field on axis due to a uniformly charged ring = $$ \frac { Q z } { 4π \epsilon (R²+ z²)^{3/2} }
$$ where z is distance of the point from the center of the ring on the axis and R us the radius of the ring.
So I tried it out by taking a patch of area da at an angle theta from the x axis and rotating it around the axis, this gives you a cone whose, locus is that of a uniformly charged ring since all of the area is at the same angle theta and the surface charge density varies with theta.

My solution :
IMG20240525152313.jpg

IMG_20240525_152345.jpg


I used integration by substitution therefore u=cos ##\theta##

IMG20240525152329.jpg

Now this answer matches with the required answer

But the suggested solution was :
1716631045699_IMG20240525152431.jpg


Now im very lost in this suggested solution and I have some questions ....
From what I gathered they have taken two spheres of opposite charges and super imposed them on each other
1) How do they know that the charge is distributed like this? That half of it is negative and the other half is positive. ..

Couldn't it be some other distribution?

2) I have derived this formula for electric field in a cavity when the cavity was in a wholely positive sphere...doesnt the half negative and half positive distribution change the formula a bit?
 
Physics news on Phys.org
  • #2
tellmesomething said:
Homework Statement: Theres a hollow sphere whose ## \sigma ## = ## \sigma_0 cos \theta ## where ##\theta ## is the polar angle/THE angle made with the x axis. Find field inside shell.
Relevant Equations: Expression of electric field on axis due to a uniformly charged ring = $$ \frac { Q z } { 4π \epsilon (R²+ z²)^{3/2} }
$$ where z is distance of the point from the center of the ring on the axis and R us the radius of the ring.

So I tried it out by taking a patch of area da at an angle theta from the x axis and rotating it around the axis, this gives you a cone whose, locus is that of a uniformly charged ring since all of the area is at the same angle theta and the surface charge density varies with theta.

My solution :View attachment 345940
View attachment 345941

I used integration by substitution therefore u=cos ##\theta##

View attachment 345942
Now this answer matches with the required answer

But the suggested solution was :View attachment 345943

Now im very lost in this suggested solution and I have some questions ....
From what I gathered they have taken two spheres of opposite charges and super imposed them on each other
1) How do they know that the charge is distributed like this? That half of it is negative and the other half is positive. ..

Couldn't it be some other distribution?

2) I have derived this formula for electric field in a cavity when the cavity was in a wholely positive sphere...doesnt the half negative and half positive distribution change the formula a bit?
Ive understood my First doubt..its merely the angle since cos theta becomes negative after 90°, the distribution becomes negative..the second doubt still somewhat remains
 
  • #3
tellmesomething said:
2) I have derived this formula for electric field in a cavity when the cavity was in a wholely positive sphere...doesnt the half negative and half positive distribution change the formula a bit?
The two superimposed spheres are both uniformly charged. The field is the sum of their fields.
 
  • Like
Likes tellmesomething
  • #4
Orodruin said:
The two superimposed spheres are both uniformly charged. The field is the sum of their fields.
Yes I figured it out. Thankyou :)
 
  • #5
Is the problem to find the field at the centre or anywhere inside the sphere?
 
  • #6
Orodruin said:
Is the problem to find the field at the centre or anywhere inside the sphere?
It is technically anywhere inside the sphere since the result is independent of distance, its uniform..
 
  • #7
tellmesomething said:
It is technically anywhere inside the sphere since the result is independent of distance, its uniform..
It is indeed, but this would have to be proven.
 
  • Like
Likes tellmesomething
  • #8
Orodruin said:
It is indeed, but this would have to be proven.
I dont follow? Would I have to prove it more formally than the solution above?
 
  • #9
tellmesomething said:
I dont follow? Would I have to prove it more formally than the solution above?
You have not shown that this is the case. It is not sufficient to just state that the field is uniform. You need to be able to argue that this is actually the case.
 
  • Like
Likes PhDeezNutz
  • #10
Orodruin said:
You have not shown that this is the case. It is not sufficient to just state that the field is uniform. You need to be able to argue that this is actually the case.
Oh. I see.. Well I dont think I know how to...
 
  • #11
tellmesomething said:
Oh. I see.. Well I dont think I know how to...
There are several possibilities depending on the level you start at. If you are familiar with vector addition and the field inside a uniformly charged sphere then you can use that. If you are more mathematically inclined you can solve Poisson’s equation to find the actual potential inside the sphere or solve the full integral for an arbitrary point.
 
  • #12
Orodruin said:
There are several possibilities depending on the level you start at. If you are familiar with vector addition and the field inside a uniformly charged sphere then you can use that. If you are more mathematically inclined you can solve Poisson’s equation to find the actual potential inside the sphere or solve the full integral for an arbitrary point.
Im familiar with the first one. Do I just have to take any other point and prove that the field will be the same there?
 
  • #13
tellmesomething said:
Im familiar with the first one. Do I just have to take any other point and prove that the field will be the same there?
Just do what they did in the suggested solution. Write down the field inside each of the uniformly charged spheres, then add them up.
 
  • #14
Orodruin said:
Just do what they did in the suggested solution. Write down the field inside each of the uniformly charged spheres, then add them up.
No that I did as I said in post #4. I thought meant an additional something after even doing the above
 
  • #15
tellmesomething said:
No that I did as I said in post #4. I thought meant an additional something after even doing the above
So if you do the vector additions of the fields proportional to ##\vec x##, then yes you get the uniformity. I am just saying it because the full argument has not been presented in the thread. It would be useful to do so both for future readers as well as for having confirmation that you got it correct.
 
  • #16
Orodruin said:
So if you do the vector additions of the fields proportional to ##\vec x##, then yes you get the uniformity. I am just saying it because the full argument has not been presented in the thread. It would be useful to do so both for future readers as well as for having confirmation that you got it correct.
Oh I see. In that case I will fair ir and post here, if you have some time please check. I would have edited it into the question but sadly I cannot edit it anymore
 
  • #17
@tellmesomething - I will advise you to take a bit of time and type out your solutions on here, using ##\rm{LaTeX}##. It is difficult to read from pictures of pages on which you have done your solution due to various reasons - one's handwriting and the lighting in the room being two of them.

Additionally, if you need to draw, there are excellent drawing softwares available, with Inkscape being my favourite. It's free. Inkspace even allows you to mark places in your drawing or write formulas in it using LaTeX.

The combined effect of typing your solutions along with the drawings is that it would be much easier for people to read and understand what you have been trying to do.
 
  • Like
Likes PhDeezNutz and tellmesomething
  • #18
brotherbobby said:
@tellmesomething - I will advise you to take a bit of time and type out your solutions on here, using ##\rm{LaTeX}##. It is difficult to read from pictures of pages on which you have done your solution due to various reasons - one's handwriting and the lighting in the room being two of them.

Additionally, if you need to draw, there are excellent drawing softwares available, with Inkscape being my favourite. It's free. Inkspace even allows you to mark places in your drawing or write formulas in it using LaTeX.

The combined effect of typing your solutions along with the drawings is that it would be much easier for people to read and understand what you have been trying to do.
Noted. Thankyou :)
 

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
868
  • Introductory Physics Homework Help
Replies
6
Views
302
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
582
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
21
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
12K
Back
Top