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

Explanation

Energy stored in the capacitor is given by
`E=1/2 CV^(2) and C=(epsi_(0)A)/(d)`, where d is the separation by between the parallel plate capacitor.
When d is decreased by 10% = 0.10
New capacitance, `C_1=(epsi_(0)A)/(0.90) =(10 epsi_(0) A)/(9)=(10)/(9)C`
New energy stored, `E_(1)=1/2. (10)/(9) CV^(2)=(10)/(9) E`
Change `=E_(1)-E`
`"% change"=(E_(1)-E)/(E) xx 100=((10)/(9)-1) xx 100=11.1%`