How to show that the electric field inside a spherical shell is zero?

In summary: Try ##\pi## instead of ##\pi/2##.I believe the only mistake you made was in your expression for the length of the blue line. See if you can show that it is ##\frac{L}{\cos \theta} d\theta## instead of ##L d\theta##.Also, make sure you have the correct limits of... integrations. Try ##\pi## instead of ##\pi/2##.
  • #1
Davidllerenav
424
14
Homework Statement
The electric field outside and an infinitesimal distance away from a
uniformly charged spherical shell, with radius R and surface charge
density σ, is given by Eq. (1.42) as σ/0. Derive this in the following
way.
(a) Slice the shell into rings (symmetrically located with respect to
the point in question), and then integrate the field contributions
from all the rings. You should obtain the incorrect result of
##\frac{\sigma}{2\epsilon_0}##.
(b) Why isn’t the result correct? Explain how to modify it to obtain
the correct result of ##\frac{\sigma}{2\epsilon_0}##. Hint: You could very well have performed
the above integral in an effort to obtain the electric
field an infinitesimal distance inside the shell, where we know
the field is zero. Does the above integration provide a good
description of what’s going on for points on the shell that are
very close to the point in question?
Relevant Equations
Coulomb's Law
WhatsApp Image 2019-09-11 at 9.32.19 PM.jpeg

Hi! I need help with this problem. I tried to do it the way you can see in the picture. I then has this:
##dE_z=dE\cdot \cos\theta## thus ##dE_z=\frac{\sigma dA}{4\pi\epsilon_0}\cos\theta=\frac{\sigma 2\pi L^2\sin\theta d\theta}{4\pi\epsilon_0 L^2}\cos\theta##.
Then I integrated and ended up with ##E=\frac{\sigma}{2\epsilon_0}\int \sin\theta\cos\theta d\theta##. The problem is that I don't know what are the limits of integrations, I first tried with ##\pi##, but I got 0. What am I doing wrong?
 
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  • #2
Davidllerenav said:
The problem is that I don't know what are the limits of integrations, I first tried with ##\pi##, but I got 0. What am I doing wrong?
If your point is at surface of sphere, the integral should be from 0 to ##\pi##/2, because sphere would have no points "higher" than your field calculation point. Integral from 0 to ##\pi## is for points inside the sphere.
 
  • #3
trurle said:
If your point is at surface of sphere, the integral should be from 0 to ##\pi##/2, because sphere would have no points "higher" than your field calculation point. Integral from 0 to ##\pi## is for points inside the sphere.
Could you please explain that a bit further? Also, I tried to compute the integraf from 0 to ##pi##/2, but I got ##\frac{1}{2}##, so the field would be ##\frac{\sigma}{4\epsilon_0}##, what am I doing wrong?
 
  • #4
Davidllerenav said:
Could you please explain that a bit further? Also, I tried to compute the integraf from 0 to ##pi##/2, but I got ##\frac{1}{2}##, so the field would be ##\frac{\sigma}{4\epsilon_0}##, what am I doing wrong?
Seems you are messing up in ##{\theta}## definition. Possibly, some of your equations still use ##{\theta}## calculated from center of sphere, not from point on top of sphere. You seems to erase old definition of ##{\theta}## on plot, but did not update equations.
 
  • #5
trurle said:
Seems you are messing up in {theta} definition. Possibly, some of your equations still use {theta} calculated from center of sphere, not from point on top of sphere. You seems to erase old definition of {theta} on plot, but did not update equations.
Maybe, but I can't see where I messed up. I used ##\phi## as the angle from the center of the ring, but I didn't use any angle from the center of the sphere. Maybe can you help me pointing out where I'm wrong?
 
  • #6
Hard to say where you go wrong cause you use an irregular spherical coordinate system with special ##\theta## and ##L##. I think the mistake is in the calculation of ##dA## as a function of ##L## and ##\theta##.

I used a typical spherical coordinate system, then used the cosine law to determine the distance of the surface element ##dA## from the point P (which u denote by ##L## in your work if I understood correctly) , and I ended up with $$\frac{\sigma}{2\epsilon_0}\frac{1}{2}\int_0^{\pi}\frac{\cos{\frac{\pi-\theta}{2}}\sin\theta}{1-cos\theta}d\theta$$. The integral is equal to 2, so the final result is $$E=\frac{\sigma}{2\epsilon_0}$$. But if I understood correctly this result is wrong since the correct result is $$E=\frac{\sigma}{\epsilon_0}$$. Where do you think it goes wrong?
 
  • #7
Delta2 said:
Hard to say where you go wrong cause you use an irregular spherical coordinate system with special ##\theta## and ##L##. I think the mistake is in the calculation of ##dA## as a function of ##L## and ##\theta##.

I used a typical spherical coordinate system, then used the cosine law to determine the distance of the surface element ##dA## from the point P (which u denote by ##L## in your work if I understood correctly) , and I ended up with $$\frac{\sigma}{2\epsilon_0}\frac{1}{2}\int_0^{\pi}\frac{\cos{\frac{\pi-\theta}{2}}\sin\theta}{1-cos\theta}d\theta$$. The integral is equal to 2, so the final result is $$E=\frac{\sigma}{2\epsilon_0}$$. But if I understood correctly this result is wrong since the correct result is $$E=\frac{\sigma}{\epsilon_0}$$. Where do you think it goes wrong?
By normal typical spherical coordinate so you mean from the center of the sphere? I'll try to do it that way then.

I don't see where the mistake may be, since we are considering the shell as a union of rings, so it should be ok.
 
  • #8
Davidllerenav said:
By normal typical spherical coordinate so you mean from the center of the sphere? I'll try to do it that way then.
Yes from center of sphere and using ##R## (the distance from the center of the sphere) and ##\theta## (the inclination or polar angle) instead of ##L## and ##\theta## you use which are totally different.
 
  • #9
Davidllerenav said:
I don't see where the mistake may be, since we are considering the shell as a union of rings, so it should be ok.
1568300726031.png


I believe the only mistake you made was in your expression for the length of the blue line. See if you can show that it is ##\frac{L}{\cos \theta} d\theta## instead of ##L d\theta##.

Also, make sure you have the correct limits of integration.
 
  • #10
Davidllerenav said:
By normal typical spherical coordinate so you mean from the center of the sphere? I'll try to do it that way then.

I don't see where the mistake may be, since we are considering the shell as a union of rings, so it should be ok.
This is what I did, I don't know if I was right at saying that the distance from the center to the point is also R, but I've made a mistake, since I didn't get the correct answer.
1568301978376-562341174.jpg
 
  • #11
TSny said:
View attachment 249540

I believe the only mistake you made was in your expression for the length of the blue line. See if you can show that it is ##\frac{L}{\cos \theta} d\theta## instead of ##L d\theta##.

Also, make sure you have the correct limits of integration.
I don't understand why it is ##\frac{L}{\cos\theta}d\theta##. I tried to show it by trigonometry, but I couldn't.
 
  • #12
what you do is almost correct, except at the point where you calculate ##\cos\alpha## where I just can't understand what you doing there.

It is easy to see that the triangle containing ##\alpha## and ##\theta## is isosceles so it is ##2\alpha={\pi-\theta}##
 
  • #13
Davidllerenav said:
I don't understand why it is ##\frac{L}{\cos\theta}d\theta##. I tried to show it by trigonometry, but I couldn't.
1568305254100.png


Consider right triangle ##acb## where the right angle is at ##c##. Note ##ac \approx L d\theta##. Show that angle ##cab## can be taken to equal ##\theta##.
 
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  • #14
Delta2 said:
what you do is almost correct, except at the point where you calculate ##\cos\alpha## where I just can't understand what you doing there.

It is easy to see that the triangle containing ##\alpha## and ##\theta## is isosceles so it is ##2\alpha={\pi-\theta}##
So indeed both so of the triangle at R? I did the correction and as you, ended up with ##E=\frac{\sigma}{2\epsilon_0}##. Why it is wrong?.
 
  • #15
TSny said:
View attachment 249546

Consider right triangle ##acb## where the right angle is at ##c##. Note ##ac \approx L d\theta##. Show that angle ##cab## can be taken to equal ##\theta##.
From that picture, I can say that ac is a proyection of ac, right? so ##ld\theta=ab\cos\alpha## thus ##ab=\frac{l}{\cos\alpha}d\theta##, where ##\alpha## is the angle ##cab##, I have no idea of how ##cab## could be equal to ##\theta##.
 
  • #16
Davidllerenav said:
I did the correction and as you, ended up with ##E=\frac{\sigma}{2\epsilon_0}##. Why it is wrong?.
What do you get if you apply Gauss's Law (in its integral form) with gaussian surface the surface of the sphere ?
 
  • #17
Delta2 said:
What do you get if you apply Gauss's Law (in its integral form) with gaussian surface the surface of the sphere ?
I got 0. ##d\phi=\vec E\cdot d\vec S=EdS\cos\alpha=EdS## since the the vector of the electric field has the same direction as the normal vector to the area element. Thus ##\phi=E\int dS = E\int_0^2pi\int_0\pi R^2\sin\theta d\theta d\varphi=2\pi R^2E\int_0^\pi\sin\theta d\theta=4\pi R^2 E##. Since the Gauss's Law is ##\phi=\int \vec E\cdot d\vec S=\frac{Q_{inside}}{\epsilon_0}##, I have ##4\pi R^2 E=\frac{Q}{\epsilon_0}## and since inside the shell the is no charge, I finally have ##4\pi R^2 E=0\Rightarrow E=0##.
 
  • #18
Yes well if you apply it with gaussian surface a sphere that has radius ##R+dr## that is infinitesimally bigger than the radius of the charged sphere ##R##, then you ll get that the electric field is ##E=\frac{Q}{4\pi(R+dr)^2 \epsilon_0}## with Q the total charge enclosed that is now ##Q=\sigma 4\pi R^2##. So substituting that in the above equation and finding the limit as ##dr \to 0## you ll finally get ##E=\frac{\sigma}{\epsilon_0}##.
 
  • #19
Delta2 said:
Yes well if you apply it with gaussian surface a sphere that has radius ##R+dr## that is infinitesimally bigger than the radius of the charged sphere ##R##, then you ll get that the electric field is ##E=\frac{Q}{4\pi(R+dr)^2 \epsilon_0}## with Q the total charge enclosed that is now ##Q=\sigma 4\pi R^2##. So substituting that in the above equation and finding the limit as ##dr \to 0## you ll finally get ##E=\frac{\sigma}{\epsilon_0}##.
Oh, that makes sense. So I don't need to modify the integral, right? For instance, what would have to change in the integral in order to have the correct answer? Since the method is correct, right?
 
  • #20
I don't know why the method with integrating Coulomb's law gives incorrect answer. I suspect it has something to do with the discontinuity of the electric field E at the surface of the sphere(inside the sphere is zero, just outside the sphere jumps at ##\frac{\sigma}{\epsilon_0}##. At the moment I can't give any hint on how to modify the integral in order to get the correct answer.
 
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  • #21
Delta2 said:
I don't know why the method with integrating Coulomb's law gives incorrect answer. I suspect it has something to do with the discontinuity of the electric field E at the surface of the sphere(inside the sphere is zero, just outside the sphere jumps at ##\frac{\sigma}{\epsilon_0}##. At the moment I can't give any hint on how to modify the integral in order to get the correct answer.
Well, that's a shame, since I think that the problem is asking me to modifiy the integral. Nothing comes to mind?
 
  • #22
well what I have in mind right now is to replace the surface charge density, with a proper continuous volume charge density so that the total charge remains the same, and then perform the integration of coulomb's law in the volume of the sphere instead of the surface. But I don't know that just doesn't look so easy. Sorry but I can't think of anything else right now.
 
  • #23
Delta2 said:
well what I have in mind right now is to replace the surface charge density, with a proper continuous volume charge density so that the total charge remains the same, and then perform the integration of coulomb's law in the volume of the sphere instead of the surface. But I don't know that just doesn't look so easy. Sorry but I can't think of anything else right now.
Anyway, thanks for you help! If you think of something, please tell me, I would like to know why the integral is not correct.
 
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  • #24
Davidllerenav said:
Anyway, thanks for you help! If you think of something, please tell me, I would like to know why the integral is not correct.
The integral calculations are correct but it doesn't give the same answer as using Gauss's law because Coulomb's law(which we used in the integral) and Gauss's law(which we used in posts #17.#18) are not equivalent in this case.

In order for Coulomb's law and Gauss's law to be equivalent the electric field must be continuously differentiable everywhere, hence continuous everywhere. This is because from what i can see in my old Greek textbook from 1999 the proof of equivalence of the two laws uses divergence theorem, and continuous differentiability is required by divergence theorem.

In this case of the charged sphere with surface charge density the electric field is not continuous everywhere, hence Coulomb's law and Gauss's law are not equivalent, and that's why they yield different results.
 
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Related to How to show that the electric field inside a spherical shell is zero?

1. How do you prove that the electric field inside a spherical shell is zero?

To prove that the electric field inside a spherical shell is zero, one can use Gauss's law. This law states that the net electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. By choosing a closed surface that lies inside the spherical shell, we can show that the net electric flux is zero, which means the enclosed charge must also be zero. Therefore, the electric field inside the spherical shell must be zero.

2. Why is the electric field inside a spherical shell zero?

The electric field inside a spherical shell is zero because of the symmetry of the charge distribution. Since the charge is uniformly distributed on the surface of the spherical shell, the electric field lines will be perpendicular to the surface at every point. This means that the electric field vectors will cancel out, resulting in a net electric field of zero inside the shell.

3. Can the electric field inside a spherical shell ever be non-zero?

No, the electric field inside a spherical shell can never be non-zero. This is because of the inverse-square law, which states that the electric field strength is inversely proportional to the square of the distance from the source. Since the distance from any point inside the spherical shell to the surface is always zero, the electric field strength at that point would be infinite. However, this is not physically possible, so the electric field inside the spherical shell must be zero.

4. How does the electric field inside a spherical shell change with the presence of an external charge?

The electric field inside a spherical shell does not change with the presence of an external charge. This is because the charge distribution on the surface of the shell remains the same, regardless of the presence of external charges. Since the electric field inside the shell is solely determined by the charge distribution on the surface, it will remain zero even with the presence of an external charge.

5. Can the electric field inside a spherical shell be affected by the material of the shell?

No, the electric field inside a spherical shell is not affected by the material of the shell. This is because the electric field is determined by the charge distribution on the surface, not the material of the shell. As long as the charge distribution remains the same, the electric field inside the shell will also remain zero, regardless of the material of the shell.

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