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I Identities of fields in Maxwell's equations

  1. Feb 21, 2017 #1

    goodphy

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    Hello.

    I would like to ask one simple question. Do we need to distinguish E-field (Electric field) in Gauss's law from those in Maxwell-Faraday equation and Ampere's circuit law? I firstly thought that E-field in Gauss's law is only for electrostatics so I need to distinguish it from E-field in time-varying Maxwell equations, If I try to do some calculation with the law. But later, I have a feeling that I may need to treat them equally; E-field in Gauss's law is also E-field in other equations. Even If I apply time-varying E-field to Gauss's law, the law gives me a correct answer; RHS of Gauss's law is zero when E-field is purely from time-varying B-field, like electromagnetic waves, in a charge free zone. The same reasoning can be applied to B-field. So every E-field (or B-field) in Maxwell's equations are the same.

    Could you tell me whether or not I'm right?
     
  2. jcsd
  3. Feb 21, 2017 #2

    Charles Link

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    Yes, the electric field ## E ## in Gauss's law ## \nabla \cdot E=\frac{\rho}{\epsilon_o } ## is the same one that appears in Faraday's ## \nabla \times E =- \frac{\partial{B}}{\partial{t}} ##. In electrostatic problems, you only need the Gauss's law equation to solve for ## E ##, and in some simpler problems with a changing magnetic field ## B ##, you only need Faraday's law to solve for ## E ## and/or the EMF ## \mathcal{E} ##. The whole set of Maxwell's equation ties the ## E ## and ## B ## together, and in the more complex cases, you may need all of the equations to solve for the ## E ## and ## B ##.
     
  4. Feb 21, 2017 #3

    goodphy

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    Thanks for replying very quick comment!

    So, E or B-fields in Maxwell's equations are all the same one, in fact.

    Could you help me a little bit more? I actually want to derive that RHS of [tex]\nabla \cdot E = \frac{\rho }{{{\varepsilon _0}}}[/tex] is zero when E-field here is E-field accompanied with time-varying B-field in a charge free zone. Could you give me some help so I can get this obvious result?
     
  5. Feb 21, 2017 #4

    Charles Link

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    When the right hand side is zero=in a charge free zone, that does not mean the ## E ## on the left hand side needs to be zero. (In the static case ## E ## will be zero, but not for cases where there is a time varying ## B ## field.) ## \nabla \cdot E=0 ## is basically a homogeneous differential equation, and there can and will be times where such a homogeneous solution results, and this equation along with the other Maxwell's equations will determine what that homogeneous solution is. As is the case with other homogeneous solutions, there are arbitrary constants and in this case even the entire solution can only be determined when the solution from the other Maxwell equations is applied. It will vary from case to case and there is no simple prescription to determine the solution. ## \\ ## Editing... For the case of ## \nabla \times E=0 ## everywhere (Faraday's law with no changing ## B ## field ), when you have ## \nabla \cdot E=0 ## everywhere as well, I believe the only solution is then that ## E=0 ## everywhere. (By standard methods, you would determine that the solution for the potential ## \Phi=constant ##, so that ## E=0 ## everywhere).
     
    Last edited: Feb 21, 2017
  6. Feb 21, 2017 #5

    Charles Link

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    @goodphy I edited my previous response. Please read the latest updated version as well.
     
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