the seperation between the plates of a charged parallel capacitor is increased.which of the following quantities will change?

The energy stored in the capacitor

The reasoning is that the voltage across the capacitor V is doubled while the charge Q remains constant, and the energy is therefore doubled: Energy = (QV)/2 = Q2/2C = (V2C) / 2 If distance is doubled then: [ (2V)2 * (C/2) / 2 ] = ( 2V2C ) / 2 = V2C Capacitance is halved due to the empirical relation C = Q / V= [ (permittivity) * A ] / d.