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Homework Help: Rate of Change of electric field

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data
    At a given instant, a 3.8 A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 1.76 cm on a side?

    2. Relevant equations

    3. The attempt at a solution
    I don't really know where to start...
    Could you use the change in length=perm of free space(I +E(sub o)*Change in E) ? I don't know if this is even an equation..
    Last edited: Mar 27, 2010
  2. jcsd
  3. Mar 27, 2010 #2
    For a parallel-plate capacitor, you can write the potential difference across the plates as

    V = E*d

    with as the potential difference, E as the electric field, and d as the separation distance. You can rearrange for E

    E = V/d

    If you look at V, we can relate it to the capacitance, C, by the definition C = Q/V. This becomes V = Q/C

    E = Q/(Cd)

    If we take the rate of chance of the left hand side and right hand side with respect to time,

    dE/dt = 1/(Cd)*dQ/dt, but dQ/dt = I = the 3.8A you were given.

    We can then replace C with the expression you attempted, C = ([tex]\epsilon[/tex]o)*(A/d) (this only holds for parallel plates!), cancel d, and we're left with an equation for dE/dt (rate of change for the electric field) with constants and your given quantities, I and the area, A.
  4. Mar 27, 2010 #3
    makes so much sense with calculus.. our prof doesn't want us using calc b/c its not advanced 1st year physics.. I didn't think about deriving the equation, just assumed there was one already set.. thanks so much!
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