How charge density of plate is changed by permittivity

[Aleksandre]

Homework Statement

Distance between plates of a parallel-plate vacuum capacitor is d. The capacitor is attached to a battery that keeps it at a voltage V. The space between plates is now filled dielectric relative permittivity epsilon. How will charge density of the plates change?

Homework Equations

sigma = charge density
epsilon₀ = The vacuum permittivity constant
E=sigma/2*epsilon₀

The Attempt at a Solution

I am not sure how to approach the problem. Theoretically, I suppose that charge density would decrease, a vacuum does not let Electric field decrease which is inversely proportional to charge. But as soon as a dielectric is used, the electric field decreases so the charge density of plate decreases. Unfortunately, I do not know which would be relevant equations to use here so that I could use charge density, voltage, d and all relative quantities. Hope you can give me some directions.

 

[Merlin3189]

For a given sized capacitor with a known dielectric, you know how much charge is stored at a given voltage, because you can calculate the capacitance from the dimensions and the dielectric.

Edit: dielectric permittivity

 

[Aleksandre]

For a given sized capacitor with a known dielectric, you know how much charge is stored at a given voltage, because you can calculate the capacitance from the dimensions and the dielectric.

Edit: dielectric permittivity

Does not capacitance formula require an area of the plate? C=epsilon*Area/distance

 

[Merlin3189]

You could you just assume an arbitrary area. Say unit area?