Electric field of two conducting concentric spheres

In summary: E \oint dA##.What is that characteristic?The characteristic is that the electric field is constant over the surface of the sphere, allowing it to be pulled out of the integral as a constant.
  • #1
MarcusNTran
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Homework Statement


Two conducting hollow spheres are are placed concentrically, the inner sphere have a radius ra = 5 cm and the outer sphere have a radius rb = 15 cm. The charge on the inner sphere is qa = 4 · 10−7 C and qb = −4 · 10−7 on the outer sphere.
(a) Use Gauss’s law to find the electric field outside the two spheres, r > rb.
(b) Use Gauss’s law to find the electric field inside the inner sphere, r < ra.
(c) Use Gauss’s law to find an expression for the electric field between the two spheres, ra ≤ r ≤ rb

Homework Equations


E*dA=Q/e0

The Attempt at a Solution



I tried to use the formula E*dA=Q/e0 but from what i got on from a and b is that q=0 in both the problems am i doing anything wrong? Q in problem a is Qa+Qb=0 and in problem b, because it is inside the sphere Q will always be 0
 
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  • #2
Hello. Welcome to PF.

Should the formula E*dA = Q/e0 have an integral symbol somewhere?

Can you describe in more detail how you are applying this formula to part (a)?
 
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  • #3
895abba6a7cf85e2b26cba289d46bcf2.png

I used this formula, simplified the formula to become E*A=Q/E the A replaced with 2*Pi*r^2 but in both problem a and b, i found out that Q becomes zero. due to if it is hollow there will be no electrical field inside the hollow object and in the other problem q becomes zero because qa and qb cancel each other out. but it does not feel right that both a and b have the same answer, so i was just wondering if i calculated Qinside right
 

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  • #4
MarcusNTran said:
View attachment 216445
I used this formula, simplified the formula to become E*A=Q/E the A replaced with 2*Pi*r^2 but in both problem a and b,
The area A is not 2πr2. What is the geometrical shape of the area A?

i found out that Q becomes zero. due to if it is hollow there will be no electrical field inside the hollow object and in the other problem q becomes zero because qa and qb cancel each other out. but it does not feel right that both a and b have the same answer, so i was just wondering if i calculated Qinside right
You didn't describe the closed surfaces that you chose for parts (a) and (b). But if you chose the proper surfaces, then, yes, Q would be zero for the charge enclosed by the surfaces for parts (a) and (b).
 
  • #5
meant 4πr^2, so the answer in a and b is E=0? I used a sylinder as the closed surface
 
  • #6
MarcusNTran said:
meant 4πr^2,
OK

so the answer in a and b is E=0?
Yes. Since you didn't describe the steps of reasoning nor your choice of surfaces for parts (a) and (b), I can't be sure that you are thinking through the problem correctly. But, your answer is correct.
 
  • #7
TSny said:
Yes. Since you didn't describe the steps of reasoning nor your choice of surfaces for parts (a) and (b), I can't be sure that you are thinking through the problem correctly. But, your answer is correct.
tbh i only used a sphererical gaussian surface mainly because it said i should use it on my textbook. and i filled in the formula, but it just seemed abit weird that the exam had the same answer in both of problems.
MarcusNTran said:
I used a sylinder as the closed surface
typo meant spherical
 
  • #8
MarcusNTran said:
tbh i only used a sphererical gaussian surface mainly because it said i should use it on my textbook. and i filled in the formula
There is a good reason for choosing a spherical shaped surface. For example, why is it that you can replace ##\oint \vec E \cdot d \vec A## by ##EA##?
 
  • #9
TSny said:
There is a good reason for choosing a spherical shaped surface. For example, why is it that you can replace
the electric field line would be perpendicular to the surface of the sphere, and since we multiply electric field and the area element and since theyre perpendicular to each other then it can be written as E*intergral dA=Q/e and the intergral dA is the area that its replaced with 4*pi*r^2, correct me if I am wrong
 
  • #10
OK. Yes, when you choose a spherical surface, ##\vec E## is perpendicular to ##d \vec A##. This allows you to replace ##\vec E \cdot d \vec A## by ##E \, dA##. In addition, there is another important characteristic of E that allows you to "pull out" E from the integral to get ##E \oint dA##.
 
  • #11
TSny said:
OK. Yes, when you choose a spherical surface, ##\vec E## is perpendicular to ##d \vec A##. This allows you to replace ##\vec E \cdot d \vec A## by ##E \, dA##. In addition, there is another important characteristic of E that allows you to "pull out" E from the integral to get ##E \oint dA##.
hmmm, what kind of important characteristic is it?
 
  • #12
MarcusNTran said:
hmmm, what kind of important characteristic is it?
What can you say about the magnitude of ##\vec E## at different points of the spherical surface?
 
  • #13
TSny said:
What can you say about the magnitude of ##\vec E## at different points of the spherical surface?
It is the total flux in that point? or it is perpendicular to the sphere?
 
  • #14
MarcusNTran said:
It is the total flux in that point? or it is perpendicular to the sphere?
It is important when applying Gauss' law to understand how symmetry is used. You are dealing with conductors of spherical shape where the charge density is spread uniformly on either the inner or outer surfaces of the conductors. When you choose a Gaussian surface of spherical shape that has its center at the center of the spherical conductors, then you can use symmetry arguments to simplify ##\oint \vec E \cdot d \vec A## over the Gaussian surface.

You have already noted that ##\vec E## is perpendicular to the Gaussian surface (i.e., parallel to ##d \vec A##) at each point of the Gaussian surface. The reason for this fact derives from the spherical symmetry of the charge distribution of the conductors. So, you are using a symmetry argument to deduce that ##\vec E## is parallel to ##d \vec A## so that you can write ##\vec E \cdot d \vec A## as ##E \, dA##.

But after doing this, you also pulled ##E## out of the integral to write ##\oint E dA## as ##E \oint dA##. What is the justification for doing this? Can you use a symmetry argument?
 
  • #15
TSny said:
But after doing this, you also pulled EEE out of the integral to write ∮EdA∮EdA\oint E dA as E∮dAE∮dAE \oint dA. What is the justification for doing this? Can you use a symmetry argument?
I moved the intergral because when we multiply the electric field to the area of element which is perpendicular to each other and the electric field is the same anywhere along the sphere then it becomes a constant that's why its become E times the intergral of dA.
 
  • #16
MarcusNTran said:
I moved the intergral because when we multiply the electric field to the area of element which is perpendicular to each other and the electric field is the same anywhere along the sphere then it becomes a constant that's why its become E times the intergral of dA.
Yes. Good. It's the symmetry of the problem combined with the symmetry of the spherical Gaussian surface that allows you to state that E is constant over the Gaussian surface.

Note that if you had chosen a Gaussian surface in the shape of a cylinder, say, then Gauss' law would still be true, but you would not be able to reduce the integral to EA.

If all of this is clear, then I think you have a good understanding of the problem.
 
  • #17
TSny said:
Yes. Good. It's the symmetry of the problem combined with the symmetry of the spherical Gaussian surface that allows you to state that E is constant over the Gaussian surface.

Note that if you had chosen a Gaussian surface in the shape of a cylinder, say, then Gauss' law would still be true, but you would not be able to reduce the integral to EA.

If all of this is clear, then I think you have a good understanding of the problem.

hmmm, I am not sure if i understand why is it that i can not reduce the intergral to EA when it comes to a sylinder? could you elaborate would help me quite a lot in the upcoming exam ^^
 
  • #18
There are two properties that ##\vec E## must have in order to reduce ##\int \vec E \cdot d \vec A## to ##E \, dA##.

(1) ##\vec E## must be parallel to ##d \vec A## so that you can write ##\vec E \cdot d \vec A = E dA##.

(2) The magnitude of ##\vec E## must be constant on the Gaussian surface so that you can pull E out of the integral.

Also note that

(3) If part of the surface is such that ##\vec E## is perpendicular to ##d \vec A##, then there is no flux through that part of the surface. So, that part of the surface does not contribute to the flux.

Property (3) does not come into play in your particular problem, but it might in a different problem.

Also, keep in mind that the Gaussian surface must pass through the point at which you are trying to find E.

Draw a picture for your spherical conductor problem and pick a point outside the outer spherical conductor where you would like to determine E. Try to choose a cylindrical Gaussian surface that passes through this point and has the properties (1), (2), or (3) given above. Convince yourself that this cannot be done. Only a spherical Gaussian surface for this problem will work.

Of course, in a different problem a cylindrical surface might be just what you need.
 
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  • #19
ahhh I finally understand ^^.Thanks a lot for the help, really appreciate you taking the time to answer. Have a nice day :)
 

Related to Electric field of two conducting concentric spheres

1. What is the electric field between two conducting concentric spheres?

The electric field between two conducting concentric spheres is zero. This is because the electric field inside a conductor is zero, and since both spheres are conductors, there is no electric field present between them.

2. How does the electric field change as the distance between the spheres is increased?

As the distance between the spheres is increased, the electric field between them decreases. This is because the electric field follows an inverse square law, meaning that it decreases as the distance between two charges increases.

3. Can the electric field between the spheres ever be non-zero?

No, the electric field between the spheres can never be non-zero as long as both spheres are conductors. If one or both of the spheres were to become charged, then there would be an electric field between them. But as long as they remain conductors, the electric field will be zero.

4. Is the electric field uniform between the spheres?

Yes, the electric field between the spheres is uniform. This means that the strength and direction of the electric field is constant at all points between the spheres.

5. How does the radius of the spheres affect the electric field between them?

The radius of the spheres does not affect the electric field between them. As long as the spheres remain concentric, the electric field will remain zero regardless of their size. However, if the spheres were to become non-concentric, then the electric field would be affected.

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