Question

If 2 capacitors are setup in parallel and charged, then disconnected from the battery and then one of the capacitors has the distance between its plates increase three-fold, how does the e-field within that capacitor respond? What about the other capacitor?

Answer 1

it is known \[E = k \frac{ q }{ r^2 }\], this implies that the distance indirectly proportional to the electric field E, if you initially had distance r and then it increases by 3r then the new electric field would \[E = k \frac{ q }{ 9r^2 }\], E decreases by \[\frac{ 1 }{ 9 }\]