Understanding parallel plate capacitors I need help with understanding this assignment. I originally though of using the equation of Eo*A/D = Q/ΔV but I don't know how to answer this without actual numbers to plug in besides the epsilon-not value. 
The parallel plate capacitor consists of two metal plates of area A separated by a constant distance d and connected to a source of electric potential difference ΔV.  One plate is connected to a high potential terminal of a voltage source (+) and the other to the low potential terminal (-).  The difference in potential (the "voltage" applied to the capacitor) is the difference between the electric potentials (voltages) of the terminals.  When a potential difference ΔV is applied to the plates, charges +Q and - Q build up on the + and - plates of the capacitor, respectively.  Although technically always zero, the charge of the capacitor is taken to be Q.

Suppose you have a parallel plate capacitor of fixed plate area A and
  fixed plate spacing d.  Explain what happens to the magnitude of the
  charge Q as the potential difference ΔV between the plates is first
  increased from zero to a maximum value, and then decreased from the
  maximum value back to zero. How does the potential change cause the
  charge to change.
Now suppose you have a parallel plate capacitor with a variable plate
  area and constant plate spacing connected to a constant source of
  potential difference, such as a battery.  (The area in the capacitor
  equation is actually the area where plates overlap to face each other.
  Variable area can be achieved by moving the plates sideways to change
  the area where they overlap.)  What happens to the charge Q as the
  plate area changes from zero to maximum value and then back to zero
  again by sliding one plate past the other? What happens to the charge
  magnitude and how does changing the area changes the charge?
Suppose you have a parallel plate capacitor of fixed plate area A
  connected to a source of fixed potential difference ΔV.  What will
  happen to charge Q if the plate spacing is increased?  What will
  happen to Q if the plate spacing is decreased?  Describe what happens
  to Q and explain how it happens physically in response to the changing
  spacing.
Explain what would happen if the plates are allowed to touch each
  other while there are equal and opposite charges on them.
How does a dielectric filling the space between the plates cause the
  capacitance to increase?  That is to say, what is it about the
  dielectric that causes the charge on the plates to increase even with
  no change in voltage, thus increasing the capacitance?

 A: You have the right equation. Since you are not given absolute values, you should just reference your answers to the original $Q$. For example, you can say for the first part "The charge will be 0 when the voltage difference is 0. As the voltage difference increases to $\Delta V$, the charge will increase (proportionally) to $Q$". No numbers were needed...
So don't "plug in numbers" - that is using equations without understanding. This assignment is about getting a feeling for the equation. What happens to $x$ when I do $y$. Look closely at the equation, and it will give you all the answers.
Except for the last one. There, you need to adjust your equation (you have another equation in your book that contains not just $\epsilon_0$, but $\epsilon_r$...)
A: The last part is possibly the most interesting in that you have think about what happens to a dielectric when it is placed in an external electric field,
or put another way; how does the movement of charges within the dielectric change the net electric field between the plates?
