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How does a faraday cage shields against internal e-fields?

  1. Jan 30, 2013 #1

    Assume we have a point charge in the center of a shell, now the inside inner wall will get an opposite charge and the outer side of the shell will have the same charge as the point charge. Since we have a charge on the out side, how isnt there any fields outside ?? how are we protected from micro oven for example ? I dont get it no matter how much I read.

    Im sorry but Im very confused, I understand how faraday cages protect against external e fields. But, How come e-fields dont escape a faraday cage when the outer shell charge exist and is directly related to the inner point charge??

  2. jcsd
  3. Jan 30, 2013 #2

    Vanadium 50

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    What makes you think that the inside of a microwave oven has a net charge?
  4. Jan 30, 2013 #3
    I was thinking generally that there is a field inside and a field on the outside too
  5. Jan 30, 2013 #4
    First understand that a microwave oven is not a Faraday cage, that would be a different discussion.

    Second, let's assume you are talking about the electrostatic case, let's ignore what happens to magnetic fields.

    Assume you have an electrically neutral Faraday cage, i.e. equal number of electrons and positive charge carriers.

    Now insert a charge, q1, into the center of the cage without touching the cage (thought experiment).

    Applying Gauss' law to a sphere that envelopes q1 but is inside the cage will show a diverging E field that is proportional to the enclosed charge q1.

    Applying Gauss' law to a bigger sphere that envelopes q1 and the cage will show exactly the same E field. The enclosed charge has not changed, the electrically neutral cage made no contribution.

    In other words the cage did not "block" the electric flux of q1.

    So what does the cage do?

    Add a test charge q2 outside of the cage. Electric flux diverging from q1/cage will cause a force on q2.

    Now move q1 around inside of the cage. Electrons within the cage will redistribute themselves. For example, if q1 is negative and you move it close to one wall of the cage, electrons on that section of the wall will shift to the other side due to ordinary electrostatic repulsion.

    Turns out that this charge redistribution that occurs within the walls of the cage leaves the field distribution *outside* of the cage identical. In other words the force experienced by the test charge q2 does not change as I move q1 around inside of the shield. This would not be true if the cage were not there, moving q1 around would alter the magnitude and direction of the force on q2 in the absence of the cage.

    This also means that the electrostatic potential outside of the cage does not depend on the position of q1 when the cage is present.

    This is the value of a Faraday cage. It maintains a constant electrostatic potential outside so long as no charges are added/subtracted from the interior (no current flowing in or out).
    Last edited: Jan 30, 2013
  6. Jan 30, 2013 #5
    The energy inside a microwave is from em waves and not electrical charges being bounced around. The em waves are reflected in the same manner as a mirror reflects ordinary light.
  7. Jan 30, 2013 #6
    Where do you think the em waves come from? Electrons are being bounced around inside of the klystron that is inside the oven.

    What do you think causes light to be reflected from a mirror?

    A microwave oven is not a Faraday cage, the "bouncing" that occurs at the perimeter of the door is accomplished using quarter wave chokes and is an entirely separate subject.

    It is a common misconception that when em waves "bounce" off of a conductor, the conductor is simply a passive barrier. In fact, the reflection is caused by the active motion of electrons within the conductor responding to the incident field.

    The OP seems to be asking about the electrostatic behavior of Faraday cage.
    Last edited: Jan 30, 2013
  8. Jan 31, 2013 #7
    Actually, the electrons are spinning around in a circle past resonant cavities inside a magnetron. There isn't a klystron tube inside any microwave I've ever seen. The electrons themselves never leave the magnetron.

    Regarding the reflection of EM waves, I think you're being overly analytic. The reflection of the wave is due to the mismatch in impedance between the conductor and free space. Yes, you can argue that the electrons in the conductor are responsible for this impedance. But it's easier to work on the macroscopic scale.

    The wavelength of a microwave oven is a little over 12 cm.
  9. Jan 31, 2013 #8
    You are right about magnetron vs. klystron (I am aware of this too, guess I was pretty tired last night).

    The OP appeared to be interested in the electrostatic behavior or Faraday cage, we are discussing the interaction of EM waves with the cage which is o.k. but may be little off his topic.

    Anyway, I have to disagree about the importance of understanding the role of electron motion, or shield currents, in EM shielding. Engineers that do not understand this are really handicapped when they try to design proper shielding. It is more common than you can imagine for engineers to struggle with seemingly counter intuitive behavior that is easily understood when considering shield currents. Let me give two examples:

    1 - I have a long cut or slit in one wall of the Faraday cage, so narrow that it is barely visible. If it were a mirror it would still reflect 99.9% of the light. If I spray water on it 99.9% of it would bounce off. If I illuminate it with EM wave, it all reflects off. Now I rotate the polarization of the EM wave by 90 degrees and it *all* passes straight through. How easy is it to explain this from a "mismatch of impedance" point of view? On the other hand lets think about the electrons. Consider two points on the wall, A and B, at the midpoint and on opposite sides of the slit, and the polarization of the incident EM wave wanting to make current flow in the direction from A to B. The only available path from A to B is the round trip out to the edge of the slit and back. The electrons are unable to redistribute themselves around this long slit in a timely manner. We have completely nullified the shielding by blocking the necessary shield currents.

    2 - Consider those holes in the door of the microwave oven. If I spray water on that material a lot would pass through. A lot of incident light passes through too (you can see your food). However virtually all of the EM wave bounces off. What happened to "The em waves are reflected in the same manner as a mirror reflects ordinary light." Real easy to understand when you consider the current flow. Electrons are free to shift around as required in reaction the incident field. Currents are free to flow in any particular direction, the time lag associated with detours to go around the holes is insignificant at intended frequency.

    I have seem numerous poor attempts to build EM shielding where the designer imagined confining EM waves to be like confining light. Lots of attention to that 1cm square hole, while ignoring the 10cm long seam between two very close surfaces.
    Last edited: Jan 31, 2013
  10. Jan 31, 2013 #9
    Indeed actually I was only referring to electrostatic fields and yes my bad a micro oven was a very bad example apparently for my question.

    yes, no matter where you move the inner q1, the outer surface charge is not gonna change, so as you said the direction and magnitude of force on q2 therefore will not change. Good so far.


    1-What about the simple outer shell charge that is constant and always there on the surface of the cage? it is the same as q1 charge, and this outer shell charge indeed as we know have a force on q2.
    2-and what if q1-charge(while centered, not touching innerwalls) is increasing and decreasing in charge or lets say alternating between positive and negative. the outside wall is directly dependent on the inner q1, so if q1(without moving) increases or reversing the charge, same happens to the outside wall and thus q2 is affected.

    So while the Faraday cage shields against any movements of q1, it doesnt shield against the q1 itself, do you see my wondering now?

    thanks a lot for all your help
  11. Jan 31, 2013 #10
    Q1 won't spontaneously change charge. That would violate conservation of mass/energy.

    And yes, the Faraday cage will not shield the charge q1 unless it's grounded.
  12. Jan 31, 2013 #11
    1 - There is no relationship between q1 and the electrostatic charge on the cage unless you bring them into contact with each other, or allow electrons to flow onto or off of the shield (by grounding it for example).

    2 - Again the electrostatic charge on the cage will only change if electrons are free to move on or off of it. It will not depend on the magnitude of polarity of q1 (unless, for example, cage is grounded). This is why, in my first post, I assumed it was electrostatically neutral to simplify the situation. If q1 somehow magically reversed polarity or changed in magnitude, this would be visible outside of a floating cage. Just imagine applying Gauss' law to a sphere that encompasses both q1 and the cage, and the electrostatic charge on the cage as a constant.

    You are right, the shield does not prevent the whole assembly from having a net E-field diverging from it due to either net charge inside the shield, on net charge on the cage.

    If you ground the cage, then you will have blocked electrostatic fields from the interior. What happens here is that the net charge inside causes electrons to flow onto or off of the entire shield to ground due to normal electrostatic forces. This leaves the potential of the cage at ground potential.

    Again, the important point is that it is the movement of charges that accomplishes all of this. The walls do not just block fields the way that a brick wall blocks basketballs.
    Last edited: Jan 31, 2013
  13. Jan 31, 2013 #12
    Im talking about an ungrounded cage, so lets see:

    Really? let q1 be negative, in the center and not touching the inner wall. The inner wall will automatically be positive and outer shell will be negative.

    Let q1 be now positive, the inner wall be negative and the outer wall will be positive. so how are q1 and the outer wall NOT related??

    if you reverse or vary the magnitude of the stationary q1 we will witness the changes as well on the outer wall. (q1 could be anything, a generator, a tesla coil, etc)
  14. Jan 31, 2013 #13
    Not only that, but the outside charges would redistribute themselves so that q1 on the inside feels no force from q2... After considering a super position of q2 and the outer surface, no net field was added inside the cage!
  15. Feb 1, 2013 #14

    It seems counter intuitive, but you have to trust Gauss' law. You are right in that electrons within the cage material will shift around in a radial direction if the internal charge changes or switches magnitude but this will not be visible externally.

    Let's take an extreme example. Three spheres with common center. Small sphere is our q1, next larger sphere is our cage, largest sphere is where we apply Gauss' law (let's call it external sphere).
    1 - q1 and sphere are electrically neutral. Charge enclosed by external sphere = 0 so no external E-field.
    2- We strip off all of the electrons from the cage and deposit them onto q1. The cage now has massive + charge. You are imagining (I think) that an external test charge would see this. It doesn't. The total charge enclosed by the external sphere is still 0, the external E-field everywhere outside is still 0.
  16. Feb 4, 2013 #15
    emi guy hmm thanx I think you perhaps captured what I was wondering, but your latest explanation in 1 and 2 sound indeed very opposite and confusing to what I thought,

    Im gonna see if I can do some testing and see, Il come back
  17. May 1, 2013 #16
    I found this post regarding farady cage very useful...I have a doubt...

    Suppose there is a spherical metallic shell with point charge q1 at the center and q2 charge outside the sphere in space ...

    Now what I have have read in this forum and on other places is that the induced charges rearrange themselves in such a way that they cancel the external electric field such that the electric field inside the shell is zero .

    Now ,when we say external electric field ...do we mean electric field due to q2 as well as q1 ?

    Last edited: May 1, 2013
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