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The questions specifically says this :

A parallel-plate capacitor initially has capacitance $C$. The distance between the plates is then doubled, with a $9.0V$ battery connected. The battery is then disconnected, and the plate area is doubled. Finally, a $20V$ battery is connected across the plates.

  • What is the new capacity?

I know that when the battery is connected and the separation is doubled, the capacitance is halved. But from then on I am not sure what to do. I tried to look it up online but had no success.

  • So my question is basically, what happens when we increase the distance separation of a capacitor without a battery? and then add a large voltage?

Thanks guys!

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  • $\begingroup$ You need to clarify what is meant by the word "capacity". If the battery is connected the voltage stays constant when you change the capacitance and when the battery is disconnected the charge stored on the capacitor stays constant as the capacitance changes. $\endgroup$ – Farcher Feb 15 '16 at 8:01
  • $\begingroup$ @Farcher Doesn't capacitance remain same and Voltage decreases when capacitor is disconnected. $\endgroup$ – Anubhav Goel Feb 15 '16 at 10:20
  • $\begingroup$ The final value of capacitance is indeed the same but what I am sure sure about is whether or not answers for the intermediate steps are being asked for. $\endgroup$ – Farcher Feb 15 '16 at 11:58
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So my question is basically, what happens when we increase the distance separation of a capacitor without a battery? and then add a large voltage?

Increasing the separation with the battery disconnected means that the capacitance of the capacitor decreases but the charge on the capacitor stays the same. Since $V=\frac Q C$ this means that the voltage across the capacitor has increased as has the energy stored in the capacitor.
This is not unreasonable as work has to be done separating the charged plates resulting in an increase in the electric potential energy of the capacitor.

However as @Anubhav_Goel has pointed out if all you need is the final answer then all you need to do is to compare the charged stored with a voltage of 9V across the capacitor with the charge stored when there is 20 V across the capacitor.

If you need answer to the intermediate steps then you do need to consider the fact that the capacitance of the capacitor does change.

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Capacitance is independent of Voltage applied and charge stored in capacitor.

$C = \epsilon ° A/d$

which is independent of Q and V.

As area and distance both are doubled , $C$ remains same.

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You first need to know that how the current flows through a capacitor. There are two parallel plates with a dielectric between them such as air etc. One plate is connected to the positive terminal of the battery while other is connected to the negative terminal of the battery. Charges from the negative terminal will accumulate on the plate connected to it. These negative charges will "push" more electrons from the plate connected to the positive terminal towards the positive terminal and thus will make the plate more positive. More will be the distance less will be the push. Less will be the distance more will be the push. Capacitance as we know is the ability to store the charge in the form electric field so more charges accumulate means that the capacitance has been increased.

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