Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook