A 12 μF parallel plate capacitor is charged to 200 V and then connected to the 7.0 μF capacitor. The plates are physically pulled apart to double the original separation. a)What is the stored energy? b)Explain the difference in stored energy between the two capacitors. c)Calculate the work performed by the outside agent pulling the plates apart

Note : We answer upto 3 parts, so i'll answer the first 3, you may post the question again mentioning the rest of the subparts that you'd like us to answer.

It's a question of Energy conservation :

Given :

C₁= Capacitance of a charged capacitor=12µF = 12 x 10^-6 F

C₂= Capacitance of the uncharged capacitor =2µF = 2 x 10-6 F

Supply voltage, V₁ = 200 V

Electrostatic energy stored in C₁ is given by,

When C₂ is connected to the circuit, the potential acquired by it is V₂.

According to the conservation of charge, initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C₂.

Now, plates are pulled apart, so the new capacitance will be the half of it, i.e. 

a) Electrostatic energy for the combination of two capacitors is given by,