charge a plate capacitor, disconnect the battery, increase the distance between the plates: capacitance decreases (C=keA/d, as d increases, C must decrease),
V increases (q=CV, q is constant, C decreases, V must increase),
electric field stays constant (E=V/d, both V and d increase proportionally),
energy stored increases (energy= 1/2 C V*V, C decreases, but V increases and squares).
The Attempt at a Solution
does it make sense that both the voltage and the energy increase as the distance between the plates increases? conceptually, I would think they would both decrease. can you help clarify? Thanks...