Keeping the voltage of the charging source constant, what would be the percentage change in the energy stored in a parallel plate capcitor, if the separation between in plates were to be decreased by `10%` ?

Explanation

Energy stored in a capacitor for a fixed voltage, `U=(1)/(2)CV^(2)`. Capacitor of parallel plate capacitor `C=(in_(0)A)/(d)`. If the separation between the plates is decreased by 10% new separation `d-(10)/(100)d=0.9d`.
`therefore`new capacitance `C'=(in_(0)A)/(0.9d)=(C)/(0.9)=(10)/(9)C`
% change in energy is `DeltaU=(C-C)/(C)xx100%=((C')/(C)-1)x100%=((10)/(9)-1)xx100%=(100)/(9)=11.1%`.