# Capacitors and Potential Difference

## 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. :rofl:)

"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?

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