Capacitors and Potential Difference

[Kaoi]

Homework Statement

(Note: This is the fourth part of a single question, but it's the only part I'm having trouble with. Don't worry, it's not as simple as Q/V. )

"What is the potential difference between a point midway between the [circular, parallel] plates [which are 1.43 x 10^-4 m apart] and a point that is 1.23 x 10^-4 m from one of the plates? Answer in units of V."

Givens: (Some of these I have gotten from the (proven correct) solutions from other parts of the question.)

\Delta V_{0} = 0.148 V
\Delta d_{1} = 7.15 \times 10^{-5} m
\Delta d_{2} = 1.23 \times 10^{-4} m
C_{0} = 5.264 \times 10^{-15} F
Q_{0} = 7.79 \times 10^{-14} C

Needed:
\Delta V_{N} = ?

Homework Equations

C = \frac{Q}{\Delta V}
\Delta V = E\Delta d = \frac{\Delta PE_{e}}{q}
PE_{electric} = \frac{1}{2}C\Delta V^{2} = \frac{Q^{2}}{2C}

The Attempt at a Solution

Alright. My problem with this question isn't so much mathematical as conceptual. If I could figure out some things, I could definitely apply some equations to solve it.

-Should I use point-charge equations for this?
-Since we're talking about voltage between points and not charges, can I use the charge capacity of the capacitor in my equations?
-Can I add up the distances between the points, or do I need to work them separately because they are on different sides of the source of the field?
-If I change the voltage, wouldn't I be unable to use my values for C and Q, since they rely on a certain voltage?

 

[chaoseverlasting]

Since in a parallel plate capacitor, the electric field is constant at all points, the potential gradient must also be a constant. Therefore, the potential at a point midway will be q/2v.