The answer depends upon how complicated you want your answer to become.  An example of a very similar question is posed in John D. Jackson's 3rd Edition of his _Classical Electrodynamics_, problem 6.14.  There are a couple of ways to approach the problem.

**First approach:**  
Assume that the current to the capacitor is given by:
$$
I\left( t \right) = I_{o} e^{-i \omega t}
$$
then the charge density, $\sigma$, on one plate at any given time is given by:
$$
\sigma \left( t \right) = \frac{ i I\left( t \right) }{ \omega \pi a^{2} }
$$
assuming the plates have a radius _a_ and separation _d_.  If we ignore the second plate for the moment (and dielectrics), we can show (using Gauss' law) that the electric field is given by:
$$
\oint dA \ \mathbf{E} \cdot \hat{\mathbf{n}} = \frac{ 1 }{ \varepsilon_{o} } \int dA \ \sigma
$$
where, in the simplest approximation (i.e., uniform $\sigma$, $\varepsilon$ = $\varepsilon_{o}$, and _d_ $\ll$ _a_), then **E** = $\sigma$/2$\varepsilon_{o}$ $\hat{\mathbf{n}}$.  A similar discussion was already posted [here](http://physics.stackexchange.com/questions/79446/electric-field-between-two-parallel-infinite-plates-of-positive-charge-and-a-gau).  

A more general result, but still simple is to consider the same idea and find the electric field on the axis of symmetry of the plate at some distance _z_ away from the plate.  Then the the above equation could be rewritten in differential form as:
$$
dE = \frac{ z dq }{ 4 \pi \varepsilon_{o} \lvert \mathbf{r} - \mathbf{r}' \rvert^{3} } \\
=  \frac{z \ \sigma \ dA}{ 4 \pi \varepsilon_{o} \left( r^{2} + z^{2} \right)^{3/2} } \\
= \frac{ 2 \pi \ z \ r \ \sigma \ dr }{ 4 \pi \varepsilon_{o} \left( r^{2} + z^{2} \right)^{3/2} }
$$
Then one could add another plate and find the superposition of the two fields.

**More general approach:**  
Assume a current is introduced at the center of the plate (i.e., r = 0), then we can write a continuity equation for the charge density as:
$$
\partial_{t} \ \sigma + \nabla \cdot \boldsymbol{\kappa} = \frac{ I\left(t\right) }{ 2 \pi r } \delta \left( r \right)
$$
where $\boldsymbol{\kappa}$ is a surface current density, $\delta \left( r \right)$ is the Dirac delta function, and $\sigma$ is a time-dependent surface charge density.  In the idealized case, Ampere's law will go to:
$$
\hat{\mathbf{n}} \times \mathbf{B} = \mu_{o} \boldsymbol{\kappa}
$$

The trick is then determining $\sigma$ and $\boldsymbol{\kappa}$ as functions of _r_, _t_, _a_, and $\omega$.  The end result is that both depend upon Bessel functions, as you might imagine from the cylindrical geometry.