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Electrostatics: Find relative permitivitty

  1. Aug 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Spherical capacitor with two linear and uniform dielectrics with relative permitivitty Ɛr1 and Ɛr2 is connected to constant voltage U. When second dielectric is removed, intensity of electric field by inner electrode is reduced by 1/3, and electric field by outer electrode is increased two times. a=1.5mm, b=6mm, c=12mm. Calculate Ɛr1 and Ɛr2.

    2. Relevant equations
    Gauss law

    3. The attempt at a solution
    In the first case (two dielectrics), Gauss law gives [tex]E_1^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2},a<r<b[/tex]
    [tex]U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)[/tex]
    In the second case (second dielectric is removed),
    [tex]E_1^{(2)}=\frac{1}{3}\times E_1^{(1)}\Rightarrow \frac{Q^{(2)}}{4\pi\epsilon_0\epsilon_{r1}r^2} = \frac{1}{3}\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r1}r^2}\Rightarrow Q^{(2)}=\frac{Q^{(1)}}{3}[/tex]
    [tex]E_2^{(2)}=2\times E_2^{(1)}[/tex]
    [tex]\frac{Q^{(2)}}{4\pi\epsilon_0 r^2} = 2\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r2} r^2}[/tex]
    From these equations, I get that [tex]\epsilon_{r2}=6[/tex]
    [tex]U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q^{(1)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)[/tex]
    [tex]U^{(1)}=U^{(2)}\Rightarrow \epsilon_{r1}=36017.1[/tex]

    In my book's solution [tex]\epsilon_{r2}=3[/tex] and [tex]\epsilon_{r1}=6[/tex]
    Could someone please check this?

    Attached Files:

  2. jcsd
  3. Aug 21, 2015 #2
    The question mentions that the electric field is reduced BY 1/3 and not to 1/3 as you have assumed.The rest of it seems right.
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