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.

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.