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If a capacitor is connected to a battery, the voltage across the capacitor will equal the battery's voltage, but this doesn't mean the capacitance is changing. The capacitance is a function of the capacitor alone (ignoring dielectrics). Right?

If I have a capacitor connected to a battery and I double the distance between the plates then the energy will be halved, right? Even though the voltage has to equal the battery's.

If I have a simple square circuit. And my capacitor comes first before my resistor (with respect to the positive plate on the battery), this shouldn't worry me, because the order wouldn't matter, right? Like, current is still in one direction only? And I could solve kirchoff's differential equation to find how my capacitor charges up?

Like ##q=C\varepsilon-C\varepsilon e^{-t/RC}## ?

And if I wanted to find the voltage across my capacitor as a function of ##t## I could divide that equation by ##C## right?

Like: ##V=\varepsilon-\varepsilon e^{-t/RC}##

Thanks.