Find the electric field inside and outside of a spherical shell superposition

In summary: I don't know what locus means, but I think that all points on the shell that have the same distance to the point would form a circle of radius ##|\vec r-\vec r'|##.Correct, the locus is a circle with radius ##|\vec r-\vec r'|## centered at ##\vec r'##. This will be important when setting up the integral to calculate the electric field.
  • #1
Davidllerenav
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Homework Statement
Given a spherical shell of radius R and charge density ##\sigma_0##, find the electric field at point A r<R and point B r>R using the superposition principle. With this results, find the analogous problem with a solid sphere with charge density ##\rho_0##
Relevant Equations
##\vec E=\frac{1}{4\pi\epsilon_0}\int \frac{\sigma}{r^2}\hat{r}##
Hi! I need help with this problem. I tried to solve it by saying that it would be the same as the field of a the spherical shell alone plus the field of a point charge -q at A or B. For the field of the spherical shell I got ##E_1=\frac{q}{a\pi\epsilon_0 R^2}=\frac{\sigma}{\epsilon_0}## and for the point charge ##E_2=\frac{-q}{4\pi\epsilon_0 r^2}##, I said that -q was the same as q, and so I could write it as ##E_2=\frac{-\sigma R^2}{\epsilon_0 r^2}##. After thaht I add them and I got ##E=\frac{\sigma}{\epsilon_0}[1-\frac{R^2}{r^2}]##. As I understand, I was meant to get ##E=0##, since at A r<R. what am I doing wrong?
 
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  • #2
Davidllerenav said:
find the electric field at point A r<R and point B r>R using the superposition principle
If that is the litteral rendering of the problem statement, you can't pretend the field of the shell is known, as in
Davidllerenav said:
saying that it would be the same as the field of a the spherical shell alone plus the field of a point charge -q at A or B
Morerover, you can't add a charge -q, because it isn't there !

I read the exercise as: add contributions from smartly chosen parts of the shell to find E for r<R and for r>R

Then in part b, the solid sphere is a sum of shells
 
  • #3
BvU said:
If that is the litteral rendering of the problem statement, you can't pretend the field of the shell is known, as in
Morerover, you can't add a charge -q, because it isn't there !

I read the exercise as: add contributions from smartly chosen parts of the shell to find E for r<R and for r>R

Then in part b, the solid sphere is a sum of shells
I see, I tried to do it by adding the -q charge because I tought I could solve it as a ploblem of a sphere with a cavity.

How should I apply the superposition principle? I was thinking of first using the charges ##dq## on the same line as the point, since those two would produce a field straight to A. all the other little charges ##dq=\sigma da## would cance in pairs, such that only the horizontal component survives. Am I correct?
 
  • #4
I would do it formally by starting with the general expression$$\vec E=\frac{1}{4\pi\epsilon_0}\int \frac{(\vec r-\vec r')}{|\vec r-\vec r'|^3}\sigma~ dA'$$Without loss of generality, you can pick the observation point at ##\vec r = z ~\hat k## but the source point must be general, ##\vec r'=R(\sin \theta '\cos\phi'~\hat i+\sin \theta '\sin\phi'~\hat j+\cos \theta '~\hat k).## Then you have to do three integrals, one for each component of the electric field. You should verify that the ##x## and ##y## components vanish as expected.
 
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  • #5
Davidllerenav said:
Am I correct?
No. Only the non-axial components cancel.

[edit] sorry, forgot to post reply this afternoon.

Kuruman's advice is good. What is the locus of points with the same ##|\vec r-\vec r'|## ?
 
  • #6
kuruman said:
I would do it formally by starting with the general expression$$\vec E=\frac{1}{4\pi\epsilon_0}\int \frac{(\vec r-\vec r')}{|\vec r-\vec r'|^3}\sigma~ dA'$$Without loss of generality, you can pick the observation point at ##\vec r = z ~\hat k## but the source point must be general, ##\vec r'=R(\sin \theta '\cos\phi'~\hat i+\sin \theta '\sin\phi'~\hat j+\cos \theta '~\hat k).## Then you have to do three integrals, one for each component of the electric field. You should verify that the ##x## and ##y## components vanish as expected.
Ok, I'll try it. I have one question, is it necessary to do it with vectors? If I know that the field should be in the radial direction, can I work with magnitudes o not?
 
  • #7
BvU said:
No. Only the non-axial components cancel.

[edit] sorry, forgot to post reply this afternoon.

Kuruman's advice is good. What is the locus of points with the same ##|\vec r-\vec r'|## ?
All points with the same vector ##|\vec r-\vec r'|## are those on the surface of the shell, right?
 
  • #8
Davidllerenav said:
All points with the same vector ##|\vec r-\vec r'|## are those on the surface of the shell, right?
All the source points are on the surface of the shell. How would you describe the locus of all points on the shell that have the same ##|\vec r-\vec r'|##?
 
  • #9
kuruman said:
All the source points are on the surface of the shell. How would you describe the locus of all points on the shell that have the same ##|\vec r-\vec r'|##?
I don't know what locus means, but I think that all points on the shell that have the same distance to the point would form a circle of radius ##|\vec r-\vec r'|##.
 
  • #10
A "locus" is the set of all points that share a common property. Example: A circle is the locus of all points that are equidistant from a single point in a two-dimensional plane. Saying they form a circle of radius ##|\vec r-\vec r'|## is not specific enough. How is this circle oriented? Remember that all you have is a shell and a point P outside (or inside) the shell. Given only these two items, how will you draw this circle on the shell?
 
  • #11
kuruman said:
A "locus" is the set of all points that share a common property. Example: A circle is the locus of all points that are equidistant from a single point in a two-dimensional plane. Saying they form a circle of radius ##|\vec r-\vec r'|## is not specific enough. How is this circle oriented? Remember that all you have is a shell and a point P outside (or inside) the shell. Given only these two items, how will you draw this circle on the shell?
By oriented, do you mean clockwise or counterclockwise? I don't know.
 
  • #12
When in doubt, make a sketch to clear up your thinking !
 
  • #13
Davidllerenav said:
By oriented, do you mean clockwise or counterclockwise? I don't know.
Along what direction (positive or negative, it doesn't matter) is the normal to the plane of the circle? The suggestion by @BvU to make a sketch is certainly helpful.

However, you can also forget about the circle of radius ##|\vec r-\vec r'|## and just do the three integrals as I suggested in #4. When you are done with the integral over ##\phi '## (do that first), you will have verified that ##E_x=E_y=0## while the integral for ##E_z## will be the same as the one you get when you consider said circle.
 

FAQ: Find the electric field inside and outside of a spherical shell superposition

1. What is the formula for calculating the electric field inside a spherical shell?

The electric field inside a spherical shell can be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the total charge of the shell, and r is the distance from the center of the shell to the point where the electric field is being measured.

2. How is the electric field inside a spherical shell affected by the presence of other charges outside the shell?

The electric field inside a spherical shell is not affected by the presence of other charges outside the shell. This is because the shell acts as a Faraday cage and the charges outside the shell do not have any influence on the electric field inside.

3. What is the formula for calculating the electric field outside a spherical shell?

The electric field outside a spherical shell can be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the total charge of the shell, and r is the distance from the center of the shell to the point where the electric field is being measured. This formula is the same as the one for calculating the electric field inside the shell.

4. How does the electric field outside a spherical shell compare to the electric field inside?

The electric field outside a spherical shell is stronger than the electric field inside. This is because the electric field inside the shell is canceled out by the charges on the inner surface of the shell, while the electric field outside the shell is not affected by these charges.

5. Can the electric field inside a spherical shell be zero?

Yes, the electric field inside a spherical shell can be zero if the total charge of the shell is zero. This means that the charges on the inner and outer surface of the shell cancel each other out, resulting in a net electric field of zero inside the shell.

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