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## Homework Statement

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same?

(a) C

(b) Q

(c) E between the plates

(d) V

## Homework Equations

[tex] C = \frac{Q}{V} [/tex]

[tex] C = \frac{\epsilon_0A}{d} [/tex]

[tex] E = \frac{\sigma}{\epsilon_0} [/tex]

## The Attempt at a Solution

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I'm doing some self study here and I'm trying to understand this problem. I'd love some thoughts on areas I'm confused on.

To start off, we are charging up this capacitor so that it holds charge Q and has a potential difference of V.

After disconnecting the battery, I know that the charge, Q, must stay the same because there's physically nowhere for the charge to go.

**So (b) remains the same.**

Now, I wanted to tackle the potential difference, V, across the plates as we pull them apart. The potential difference between the two plates is equal to the work needed per unit charge in moving from one plate to the other. That is, if I physically move a charge from one plate to another, the work I do (per unit charge) is equal to the potential difference. This work will increase (because we're moving over a longer distance) if we increase the spacing on the plates.

**And that's why the potential difference will increase as we increase the space between the plates. So (d) increases.**

The electric field, E, is given by the equation above, and is constant so it will remain the same.

**(c) remains the same. CONFUSION HERE:**I don't have a good intuition on why it is the electric field remains the same when the potential difference increases. Any thoughts here?

And finally, the capacitance. I think I have a hard time reconciling the geometry relationship for capacitance with the formula that relates it to charge/voltage. From the formulas above, we see that if the potential difference increases, that the capacity decreases.

**CONFUSION HERE:**But I always see this as the capacity to store charge and if the capacitance decreased, then its capacity to hold charge has decreased and so it should have less charge on it. But the Q can't go anywhere so I'm unclear as to how the capacitance decreases.

Any insight into the last two paragraphs would be great.