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Effect of electric field on a dielectric

  1. Aug 19, 2009 #1
    friends,
    can anyone explain me the effect of electric field on a dielectric(non- conducting) medium
     
  2. jcsd
  3. Aug 19, 2009 #2
    First, let's have a look at the effect an external field creates on an otherwise spherical atom. When the atom is placed in the external field, the positive charges tend to move along the direction of the field while negative electron gas cloud in the opposite direction. If the field is strong enough it will break the atom with the electrons ripped off. (That is ionization). But if the field is not too large, there is an effective separation of positive and negative charges on the atom which leads to the formation of a dipole. For small fields, it is found that the induced dipole [tex]\vec{p}=\alpha \vec{E}[/tex], where [tex]\alpha[/tex] is the atomic polarizability.

    Coming to a dielectric, when it is placed in an external field little dipoles are created which essentially point in the same direction. So if we consider a homogeneous dielectric, the positive charge of one dipole coincides with the negative one of the next one and effectively cancels it. This leads to the cancellation of all the charges within the bulk leaving out charges sticking on a very thin layer on the surface. The surface whose (outward) normal is opposite the field gets a negative charge while that one having the normal in the direction of the field gets a positive charge. These charge then create a field of their own which runs in the direction opposite that of the external field. The total field is the sum of the two fields: [tex]\vec{E}_\mathrm{tot}=\vec{E}_\mathrm{ext}+\vec{E}_\mathrm{d}[/tex]. Since the two fields are opposite inside the dielectric, the resultant field is less inside the dielectric.
    However, these are the consequences. The answer to your question should be: An electric field polarizes a dielectric.
     
  4. Aug 19, 2009 #3

    drizzle

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    ...you mean:
    [tex]\vec{E}_\mathrm{tot}=\vec{E}_\mathrm{ext}-\vec{E}_\mathrm{d}[/tex]
     
  5. Aug 19, 2009 #4
    no drizzle. you add them vectorically. the effect will be to subtract their magnitude since the two vectors are pointing in opposite directions.
     
  6. Aug 19, 2009 #5

    drizzle

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    I’m quite convenient with the sign of the vector been represented in the equation [it symbolizes a direction], it also could be written this way;

    [tex]\vec{E}_\mathrm{tot}=\vec{E}_\mathrm{ext}+(-\vec{E}_\mathrm{d})[/tex]
     
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