Rate of Change of electric field

[sdoyle1]

Homework Statement

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?

Homework Equations

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..

 

[Patrick_Nth]

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 = (\epsilono)*(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.

 

[sdoyle1]

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!