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Capacitors in series with dielectrics

  1. Jul 26, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    A ##240 nF## capacitor and a ##4.0 \mu F## capacitor are connected in series with a ##240 V## potential difference.

    The ##4.0 \mu F## capacitor has a parallel plate configuration, with a surface area of ##110 cm^2## and a dielectric material between the plates. The dielectric constant is ##\kappa = 8##.

    1. Calculate the charge stored in the ##4.0 \mu F## capacitor.

    2. Calculate the potential across the ##4.0 \mu F## capacitor and the potential across the ##240 nF## capacitor.

    3. Find the spacing between the plates of the ##4.0 \mu F## capacitor.

    4. Find the magnitude of the electric field in the ##4.0 \mu F## capacitor.

    5. Calculate the energy density in the ##4.0 \mu F## capacitor.

    2. Relevant equations

    ##C_1 = 4.0 \mu F = 4.0 \times 10^{-6} F##
    ##C_2 = 240 nF = 240 \times 10^{-9} F##

    ##V = 240 V##
    ##\kappa = 8##

    ##q = CV##
    ##C = \epsilon_0 L##, where ##L## has the dimension of length.

    Screenshot of how I figure the circuit looks: http://gyazo.com/6c138b176e22538339aa4ab44b899b50

    For series capacitors: ##C_{eq} = (\sum_{i=1}^{n} \frac{1}{C_i})^{-1}##

    3. The attempt at a solution

    1. I'm slightly confused about how to factor in the dielectric material.

    A potential is maintained across the circuit and a dielectric is filling the space between the plates of ##C_1##. So the effect of the dielectric is to increase the charge ##q_1## on ##C_1##, by increasing ##C_1## by factor of ##\kappa##?

    Using this logic, I could find an equivalent capacitance ##C_{12}## given by:

    ##C_{12} = ( \frac{1}{\kappa C_1} + \frac{1}{C_2} )^{-1}##

    Then using ##q_{12} = C_{12} V_{12}##, I could find the charge across ##C_{12}##.

    Then using the fact that capacitors in series all have the same charge, we could write ##q_{12} = q_1 = q_2##.

    Does this seem reasonable?
     
  2. jcsd
  3. Jul 26, 2014 #2

    vela

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    No, you're given ##C_1##. That's the capacitance with the dielectric already factored in.

    You need the dielectric constant to answer 3-5, which have to do with the internals of the capacitor.
     
    Last edited: Jul 26, 2014
  4. Jul 26, 2014 #3

    Zondrina

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    Ah I was a bit confused about that. Thank you for you input.

    1. An equivalent capacitor is given by:

    ##C_{12} = ( \frac{1}{C_1} + \frac{1}{C_2} )^{-1} = ( \frac{1}{4.0 \times 10^{-6} F} + \frac{1}{240 \times 10^{-9} F} )^{-1} = 2.2642 \times 10^{-7} = 2.3 \times 10^{-7} F##

    The charge on this capacitor is given by:

    ##q_{12} = C_{12} V_{12} = (2.2642 \times 10^{-7} F) (240 V) = 0.000054 = 5.4 \times 10^{-5} C##

    Since the capacitors are in series ##q_1 = q_2 = q_{12}##.
     
    Last edited: Jul 26, 2014
  5. Jul 26, 2014 #4

    Zondrina

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    Continuing the work here:

    2. To obtain the potential across ##C_1##, simply use:

    ##q_1 = C_1 V_1 \Rightarrow V_1 = \frac{q_1}{C_1} = \frac{5.4 \times 10^{-5} C}{4.0 \times 10^{-6} F} = 13.5 V = 13 V##

    To obtain the potential across ##C_2##:

    ##q_2 = C_2 V_2 \Rightarrow V_2 = \frac{q_2}{C_2} = \frac{5.4 \times 10^{-5} C}{240 \times 10^{-9} F} = 225 V = 230 V##

    My results seem to be slightly off, not sure why. They should add up to ##240 V##.


    3. To find the spacing between the plates of ##C_1##, simply use the general equation for a capacitor with a dielectric:

    ##C_1 = \kappa \epsilon_0 L = \kappa \epsilon_0 \frac{A_1}{d_1}##

    Hence: ##d_1 = \kappa \epsilon_0 \frac{A_1}{C_1} = (8)(8.85 \times 10^{-12} \frac{F}{m}) \frac{0.011 m^2}{4.0 \times 10^{-6} F} = 1.947 \times 10^{-7} m = 1.9 \times 10^{-7} m##


    4. Using Gauss' law with a dielectric and noting the electric field is constant between the plates:

    ##q_1 = \epsilon_0 \oint \kappa \vec{E} \cdot d \vec{A} = \kappa \epsilon_0 \oint \vec{E} \cdot d \vec{A} = \kappa \epsilon_0 E \oint dA## (Since ##\vec{E} \cdot d \vec{A} = E cos(0°) dA##).

    The closed surface integral now gives the surface area of the plates, so we write:

    ##q_1 = \kappa \epsilon_0 E A \Rightarrow E = \frac{q_1}{\kappa \epsilon_0 A} = \frac{5.4 \times 10^{-5} C}{(8) (8.85 \times 10^{-12} \frac{F}{m}) (0.011 m^2)} = 6.934 \times 10^{7} \frac{V}{m} = 6.9 \times 10^{7} \frac{V}{m}##


    5. To find the energy density of ##C_1##, simply use:

    ##u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} (8.85 \times 10^{-12} \frac{F}{m}) (6.934 \times 10^{7} \frac{V}{m})^2 = 21275.557 \frac{J}{m^3} = 21000 \frac{J}{m^3}##



    Do these look good?
     
    Last edited: Jul 26, 2014
  6. Jul 27, 2014 #5

    vela

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    It's just rounding. To the two sig figs you kept, the sum is consistent with 240 V.

    You could also use ##V=Ed## since the electric field is uniform. It's a bit simpler, though your calculation is fine too.

    You need to account for the dielectric medium here.
     
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