# Capacitors and Voltage

Imagine if I have a capacitor, connected to a battery of $$12 V$$.

After charging the capacitor, I increase the plate separation. Then my capacitance decreases, right?

But charge should be conserved, which means that since $$C=\dfrac{q}{V}\space$$ my voltage increases. But is that possible since I am only connected to a battery of $$12 V$$.

If I assume, that my voltage is $$12 V$$ (constant) then the charge stored should decrease, but then wouldn't that violate conservation of charge?

I am confused. I don't think I understand this concept really well.

Any help would be appreciated.

• Incase it is disconnected from the battery charge is conserved and voltage can vary.. But if you change the plate separation while still connected to the battery, voltage is conserved but charge can vary May 10, 2015 at 3:33
• @slhulk That is an answer, not a comment.
– rob
May 10, 2015 at 3:33
• @rob Is that remark intended to encourage posting it as an answer? Just curious. May 10, 2015 at 3:50
• @StanShunpike Yes, it is. The distinction between comments and answers is discussed in the commenting guidelines.
– rob
May 10, 2015 at 4:27
• Hmm...after reading that, I agree. I now see why you make this comment. May 10, 2015 at 4:29

If you have an isolated capacitor, so that there is no conducting path for charge to flow from one plate to the other, then the charge on the plates will be conserved as you change the geometry. Since $$Q = CV,$$ a drop in the capacitance $C$ is matched by an increase in the potential $V$. Note that the stored energy $U$ in the capacitor, $$U = \frac12 CV^2,$$ increases as you pull the plates apart; this happens because the plates are electrically attracted to each other and pulling them apart takes work.
If your capacitor is connected to some circuit, then in general $Q$ will not be conserved. For example, if a capacitor is connected to a battery, charge will flow through the battery to maintain the battery's preferred potential difference across the terminals.