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Electric field of Capacitors

  1. Jul 16, 2009 #1
    1. The problem statement, all variables and given/known data
    An electric field of a capacitor is defined by the following equation,
    [tex]\vec{E} = \frac{\rho_{s} \hat{a_n}}{\epsilon} = \frac{Q\hat{a_n}}{A\epsilon_r\epsilon_0}[/tex]
    where [tex]\vec{E} = (\frac{\rho_s}{\epsilon})\hat{a_n}[/tex]


    Question
    I understand the first equality, and reviewing my physics book, I think I understand the derivation. But I was wondering if [tex]\epsilon = \epsilon_r\epsilon_0[/tex], and a quick explanation why.


    Thanks,


    JL
     
  2. jcsd
  3. Jul 16, 2009 #2

    rl.bhat

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    Homework Helper

    Here εο is the permittivity of the vacuum and εr is the relative permittivity of the medium which is introduces between the plates. It is also called as the dielectric constant.
     
  4. Jul 16, 2009 #3
    Actually, do you mind explaining why it was necessary to break the permittivity into those components? And why the product is equivalent to the permittivity?

    Thanks again,


    JL
     
  5. Jul 16, 2009 #4

    rl.bhat

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    Homework Helper

    Permittivity ε is the property of the space. If you introduce anything between the plates of the capacity, the electric field will decrease due to polarization.. Consequently the potential difference between the plates will decrease and hence the capacitance will increase.
    The ratio of capacity with medium and capacity without medium is called relative permittivity εr. εr = Cm / Co = (εA/d)/(εoA/d) = ε/εr
     
    Last edited: Jul 16, 2009
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