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.
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.
