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Electric field in non conducting shell

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Q . Is the electric field zero inside a uniformly charged non conducting shell in the presence of an external charge ?

    2. Relevant equations



    3. The attempt at a solution

    If the external charge had not been there ,then I am sure that the electric field inside at any point within the shell would be zero .Again,if it had been conducting shell,then too the electric field would have been zero even in the presence of an external charge.

    But I think,the electric field will have a non zero value in the non conducting shell in the presence of an electric charge.

    Could someone help me in clearing the doubt ?

    Thanks
     
  2. jcsd
  3. Mar 17, 2014 #2

    TSny

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    You might be right. But can you explain why you think the field is nonzero?
     
  4. Mar 17, 2014 #3
    Hello TSny

    In case of a conducting shell,when there is a nearby external charge( external electric field) ,the charges rearrange themselves in such a a manner so as to cancel the external electric field ,both within the material of conducting metal and the inside volume of the shell.

    In case of non conducting shell,the charges do not have a chance to rearrange as they are fixed .

    But then we have the shell theorem " The electric field within a uniformly charged shell whether it is conducting or non conducting is zero " .

    Is the Shell Theorem applicable only in the absence of external electric field ?

    I am not sure about the reasoning .Could you please reflect more on this ?
     
  5. Mar 17, 2014 #4

    TSny

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    All the above sounds very good. (The charges in a nonconductor can move a little due to polarization of the molecules, but that's not very relevant here.)

    The key is the principle of superposition. The net field at a point inside the shell will be the sum of the fields from the charges of the shell plus the field of the point charge. The shell theorem tells you the field due to the charge on the shell (assuming you can treat the charge as uniformly spread on the shell).
     
    Last edited: Mar 18, 2014
  6. Mar 18, 2014 #5
    Thanks :smile:

    Q.1 A rubber balloon is given a charge Q distributed uniformly over its surface.Is the field inside the balloon zero everywhere if the balloon does not have a spherical surface ?

    Q.2 A spherical shell made of plastic,contains a charge Q distributed uniformly over its surface.What is the electric field inside the shell,if the shell is hammered to deshape it without altering the charge ?

    My reasoning - In both the questions the field will be non -zero .Considering a gaussian spherical surface within the volume .∫E.dS = 0 i.e flux will be zero .But we will not be able to take E out of the integral because of non symmetry . Hence even though flux is zero ,but electric field will be non zero .

    Is the reasoning alright ?
     
  7. Mar 18, 2014 #6

    TSny

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    Sounds good!
     
  8. Mar 18, 2014 #7
    Thanks...

    Suppose we have a uniformly charged conducting shell .There is a point charge q at its center .Now an external charge Q is placed near the shell .The charge distribution on the surface of the shell will change such that the net electric field inside the shell will remain zero .The external charge Q will induce negative charge on the nearer side of the shell with positive charge on the farther side .The electric field due to Q will not be able to penetrate the shell .

    There are two things

    1) The net electric field is zero inside the shell,hence no net force on q .
    2) The external charge Q exerts force on q ,but how ?.There is no electric field inside the shell .So how does the external charge influences the inside charge despite no electric field inside the shell?

    Could you reflect on this ?
     
    Last edited: Mar 18, 2014
  9. Mar 18, 2014 #8

    TSny

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    By "net" field here, I guess you mean the superposition of the field due to the charges on the shell and the charge Q, but not including the field of the charge q. If you include the field of q, then the net field will not be zero at points inside the hollow region of the shell.

    Phrases such as "the electric field due to Q will not be able to penetrate the shell" are a little "loose". By the superposition principle, you can just as rightly say that the field due to Q is unaffected by the presence of the conducting shell. So, the field due to Q has no problem penetrating the shell and exerting a force on q inside the shell. But, the charges on the shell also produce electric field such that the superposition of the field of Q and the field of the charge on the shell add to zero inside the shell. All that matters is the net field at the location of q. In other words, you could say that Q does exert a force on q, but the charges on the shell exert an opposite force on q.

    1) Right. The net field due to the charge of the shell and the charge Q is zero at the location of q. So, q experiences no force.

    2) Q creates an electric field at the location of q, but the charges on the shell create a field at q that cancels the field from Q.

    The external charge Q is not able to influence q because as soon as you move Q, the charges on the shell redistribute in such a way as to cancel the effect of Q on q.
     
  10. Mar 18, 2014 #9
    Scintillating explanation...:thumbs:

    The clarity in your thoughts is simply superb.

    Thank you so much :smile: .I am really really grateful to you for explaining the concepts.
     
  11. Jun 28, 2014 #10
    Hello TSny

    There is something about superposition principle and electric fields which has been bothering me for a while .Kindly give your views .

    Suppose the question is :A conducting sphere of radius R has a charge Q. A particle carrying a charge q is placed a distance 2R from the sphere. Find the potential at point A located a distance R/2 from the center of the sphere on the line connecting the center of the sphere and particle q.

    Solution 1. - Potential at the center of the sphere = kQ/R+kq/(3R) .Since the sphere is equipotential ∴ Potential at A = kQ/R+kq/(3R)

    Solution 2. - Potential at A = Potential due to sphere + Potential due to point charge = kQ/R+2kq/(5R)

    I know solution 1 is correct .But what is wrong with solution 2 ?
     

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  12. Jun 28, 2014 #11

    TSny

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    Hello, Tanya.

    How did you conclude that the potential at A due to the charge on the sphere is kQ/R?
     
  13. Jun 28, 2014 #12
    I know it is wrong . I am just trying to find the flaw in the logic.

    But here is the reasoning - By superposition principle ,the potential at the center due to sphere is kQ/R .Now since the sphere is equipotential ∴ Potential at A = Potential at center = kQ/R
     
  14. Jun 29, 2014 #13

    TSny

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    The surface and interior volume of the sphere is an equipotential region due to the combined effect of the nonuniformly distributed charge Q on the sphere and the external point charge q.

    The charge Q on the sphere, by itself, does not create an equipotential on the sphere or within the sphere.
     
  15. Jun 29, 2014 #14
    Right . That clears up the confusion :)

    Case 1) Suppose there is a non conducting uniformly charged shell of radius R having charge Q .

    If we consider a Gaussian spherical surface of radius r ,r<R then ∫E.ds = 0 and by symmetry E is same at distance r , and hence ∫E.ds = E∫ds =0 → E=0 in the shell .

    Case 2) Suppose there is a non conducting uniformly charged shell of radius R having charge Q and a point charge q outside the shell .

    If we consider a Gaussian spherical surface of radius r ,r<R then ∫E.ds = 0 but now due to the presence of external point charge ,the symmetry no longer exists and hence E cannot be taken out of the integral .So E≠ 0 in the shell .

    Is the reasoning correct ?
     
  16. Jun 29, 2014 #15

    ehild

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    Yes, the net flux can be zero when some electric field line go inward, others outward.

    ehild
     
  17. Jun 29, 2014 #16
    I think that it is indeed correct, and you can also imagine the field lines. The symmetry will break due to the charge distribution.
    However, finding the actual charge distribution in case 2 is rather difficult I reckon...

    If it was a conducting shell, then it'd act as Faraday's cage, where symmetry will be possible I think :)
     
  18. Jun 3, 2015 #17
    Hi TSny,

    Please have a look at the attached figure.

    There is a metallic spherical shell of radius R and having charge Q, a point charge q inside the shell at a distance R/2 from center O, and a point charge q1 at a distance 2R from the center.I would like to find net electric field at the center O.

    The Electric field would be due to the following charges :

    a) Charge q (towards right).
    b) Charge q1 (towards left)
    c) Induced charge –q on inside surface of shell .
    d) Charge Q+q on outer surface of shell .

    Now,if charge q was not present inside the shell then EF at the center would have been 0 ,but that is not the case here.

    1) Is electrostatic shielding present in this case ? I do not think so , as charge q is present inside the shell .
    2) How do I find EF due to c) and d) ?
    3) Is it possible to calculate net EF at the center in this case ?

    Please give your views .
     

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  19. Jun 3, 2015 #18

    TSny

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    There will be a field inside the sphere due to the charge q and the induced charge on the inner surface of the sphere. However, the sphere does shield the inside from the effects of the outside charge ##q_1##. Changing the magnitude or location of ##q_1## will not affect the net field inside the sphere.

    This can be done using the method of images.

    The electric field inside the sphere will be due to the charge ##q## and the charge induced on the inner surface of the sphere. The charge on the outer surface and the charge ##q_1## together produce no field inside the sphere.

    When finding the field inside the sphere, the contribution to the field due to the induced charge on the inner surface is equivalent to the field of a point, image charge ##q'## located outside the sphere. Standard texts explain how to determine the magnitude and location of the image charge ##q'##. The net field at a point inside the sphere will be just the superposition of the fields of ##q## and ##q'##.
     
  20. Jun 3, 2015 #19
    Ok . I will look up method of images . For now I do not understand how external charge q1 and charge on outer surface of shell produce no field inside shell .

    But from post #8 [Q played the role of q1]

    Didn't you explain in post#8 that external charges do produce electric field inside ? It is just that the net electric field inside shell is zero (if no charge is present inside )due to property of conductors .

    Could you reflect more on how external charge q1 and charge on outer surface of shell produce no field inside shell ?

    Thank you very much .
     
    Last edited: Jun 3, 2015
  21. Jun 3, 2015 #20

    TSny

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    This is semantics to some extent.

    By the principle of superposition, you can think of any point charge as producing a point-charge electric field that "exits" and is unaffected by the presence of any other charges. But what matters is the net field due to adding the individual fields of all charges.

    If you have equal point charges on the corners of a square, then the net field at the center of the square is of course zero. If you put a test charge at the center, it feels no force. You can think of this as resulting from a force from each of the corner charges and then adding the forces to get zero. Or, you can think of it as the four fields "cancelling each other out" to produce no field at the center and therefore the test charge feels no force. There is no way to distinguish these two ways of thinking about it.

    The inside of the spherical conductor is "shielded" from the effects of an outside charge Q. Does this mean that "the field of Q does not penetrate into the interior of the sphere"? I prefer to think of the point-charge field of Q existing everywhere. But the induced charges on the sphere also produce electric fields. The field of Q and the fields of the induced charges amazingly add to zero at every point inside the sphere! And, this is also true for arbitrary shaped conductors. It's a consequence of the fact that the field of a point charge is inversely proportional to the square of the distance and that charges are free to move in a conductor.
     
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