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Voltage, capacitance and EM energy of a parallel plate capacitor

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data

    This is an EM waves class, but we are given the following problem, which seems deceptively simple. I'd like some feedback to see if this is done correctly.

    2. Relevant equations

    Consider a parallel plate capacitor connected with a resistor with resistance R, as shown in the figure (it's a simple circuit of a resistor R in series with a switch and parallel plate capacitor, I can recreate if necessary; there is no source). Suppose the plate area is A, the distance between the two plates is d, and between the two plates is a uniform, lossless dielectric material with permittivity ε and permeability μ0. Assume the plate dimension is much larger than the distance d, so you can assume the E field is uniform between the plates and vanishes outside.

    A) Calculate the capacitance. Assume the initial voltage across the cap is V0, and the switch is closed at time t=0. (Voltage polarity is given in the figure.) Calculate the voltage across the capacitor as a function of time. Also calculate the power consumption of the resistor as a function of time.

    B) Calculate the total EM energy stored within the cap as a function of time. What conclusion can you draw from your calculations?

    3. The attempt at a solution

    A) [itex]V = \frac{Qd}{\varepsilon A}[/itex], where Q is the charge on the plate. As C = Q/V, I have
    [tex]C= \frac{\varepsilon A}{d}[/tex]
    And
    [tex]V = V_0 e^{-\frac{t}{RC}} = V_0 e^{-\frac{td}{R\varepsilon A}}[/tex]
    Finally, for power,
    [tex]P = \frac{V^2}{R} = \frac{V_0^2}{R}e^{-2 \frac{td}{R\varepsilon A}}[/tex]
    B) [tex] u_E = \frac{1}{2} \varepsilon E^2
    = \frac{1}{2} \varepsilon \left(\frac{V_0}{d} e^{-\frac{td}{R \varepsilon A}}\right)^2
    = \frac{1}{2} \varepsilon \left(\frac{V_0^2}{d^2}e^{-2\frac{td}{R \varepsilon A}}\right)[/tex]

    And the 'conclusion' I would reach is that the energy is entirely stored within the electric field, and the permeability does not affect the parameters of the circuit.

    Is all of this accurate or have I made any errors?

    (Sorry if this should be in the basic board, please feel free to move it there if so.)
     
    Last edited by a moderator: Sep 28, 2012
  2. jcsd
  3. Sep 28, 2012 #2
    Even more baffling, the next question set asks

    "This time, assume we are charging the capacitor with a constant current of I, starting at time t=0.

    A) Calculate the electric field within the capacitor as a function of time."

    I've done some looking around and can't seem to find nil on a constant current charging a capacitor, beyond that it would induce infinite voltage. I'm not even sure where to start on this one.
     
  4. Sep 28, 2012 #3

    vela

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    In part B, you've only calculated the energy density, not the energy stored in the capacitor as a whole.

    For the next question, think about how are Q and I related.
     
  5. Sep 30, 2012 #4
    Hey Vela, thanks for all your help. So so far I've got it down to the last part and I think I'm doing okay. The question page is attached.

    A) As current is constant, and Q = I*t, the charge on the plates will also increase at a constant rate. Thus, as

    E = V/d = [itex]\frac{Q}{εA}[/itex] = [itex]\frac{It}{εA}[/itex]

    B) uE = [itex]\frac{1}{2}[/itex]εE2 = [itex]\frac{1}{2}[/itex]ε[itex]\frac{It}{εA}[/itex]2 = [itex]\frac{1}{2ε}[/itex][itex]\frac{It}{A}[/itex]2

    This is the energy density, so I would just multiple this by the volume of the dielectric (d*A). Additionally, I would assume the current source has to supply this much energy, so wouldn't the energy provided by the current source be the same value ([itex]\frac{d}{2Aε}[/itex](It)2 ?

    C) As far as I've gotten with this is the analysis of Ampere's law on the left, with

    [itex]\oint[/itex]H[itex]\bullet[/itex]dL = Hρ02π = I

    or

    H =[itex]\frac{I}{ρ02π}[/itex]

    Not really sure how to progress form here or what I'm looking for. Any advice would be appreciated.
     

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  6. Oct 1, 2012 #5
    Can anyone shed a little light on this last problem?
     
  7. Oct 1, 2012 #6

    TSny

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    You've evaluated the left hand side of Ampere's law. I think you're asked to evaluate the right hand side and see if it agrees with the left hand side.
     
  8. Oct 3, 2012 #7
    Okay, so so far gotten to the point where for the left hand side of Ampere's law, I have

    ∮H∙dL = Hρ02π = I

    For the right side,

    We have D = εE = ε[itex]\frac{It}{εA}[/itex] = [itex]\frac{It}{A}[/itex], all in the -z direction,

    Thus, [itex]\frac{\partial}{\partial t}[/itex][itex]\frac{It}{A}[/itex] = [itex]\frac{I}{A}[/itex], again in the -z direction

    So, going back, we have

    ∫∫[[itex]\frac{\partial D}{\partial t}[/itex] + J][itex]\bullet[/itex] dS,
    J = 0, dS = ρdρd∅, thus

    ∫∫[[itex]\frac{\partial D}{\partial t}[/itex] + J][itex]\bullet[/itex] dS = ∫∫[itex]\frac{I}{A}[/itex]ρdρd∅

    Integrating from ρ = 0 to ρ0 and ∅ = 0 to 2π, this gives [EDIT: I made a slight error here and have corrected it]

    = [itex]\frac{1}{2}[/itex][itex]\frac{2πI}{A}[/itex]ρ02 = [itex]\frac{πI}{A}[/itex]ρ02

    and I really don't know what I'm supposed to do with this, because technically right now I have

    Hρ02π = [itex]\frac{πI}{A}[/itex]ρ02

    This kind of makes sense since in the RHS we have I * [itex]\frac{π}{A}[/itex]ρ02, with πρ02 of course being area. The only thing that throws me off is that A, the area of the plate, is not the same as the above integration area, but since A is encompassed in that, can I cancel them and just get I? Because that would make sense, reducing to

    Hρ02π = I

    Which is the result I would expect from Ampere's law.

    Am I accurate?
     
    Last edited: Oct 3, 2012
  9. Oct 3, 2012 #8

    TSny

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    Note that ∫dS is just the area A, where A = [itex]\pi\rho_o^2[/itex] [EDIT: This is wrong, please see my following post. I should have said that ∫dS gives just the area of the capacitor plate (not [itex]\pi\rho_o^2[/itex]) because D = 0 outside the capacitor.]
     
    Last edited: Oct 3, 2012
  10. Oct 3, 2012 #9

    TSny

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    Sorry, my previous reply is bogus. You can assume the ideal capacitor has D = 0 outside of the region between the plates. So, when you integrate over the area enclosed by the path of integration, the only nonzero contribution is over the area A of the plates. So, the area from the integration will cancel the A in (I/A).
     
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