a vacuum filled parallel plate capacitor with a capacitance of20 mf is connected

Question

a vacuum filled parallel plate capacitor with a capacitance of20 mf is connected to a 150 V source. A) what is the charge stored on the capacitor? B) what is the energy stored in the capacitor? C) if the battery is left connected to the capacitor and anylon dielectric K=4.3 is inserted between the plates, what is theenergy now stored, if different from before?
Here is my problem. I don't understand the dielectricmess.... i am given two sets of formulas... part 1. c=kco v=vo/k E=Eo/k and U=u0/k
part 2 q=kqo v=vqo    v=vo and u =kuo

so.. i don't know when to use which formula.. i think one setis for connected batteries and one isn't. any help would beappreciated. cheers a vacuum filled parallel plate capacitor with a capacitance of20 mf is connected to a 150 V source. A) what is the charge stored on the capacitor? B) what is the energy stored in the capacitor? C) if the battery is left connected to the capacitor and anylon dielectric K=4.3 is inserted between the plates, what is theenergy now stored, if different from before?
Here is my problem. I don't understand the dielectricmess.... i am given two sets of formulas... part 1. c=kco v=vo/k E=Eo/k and U=u0/k
part 2 q=kqo v=vqo    v=vo and u =kuo

so.. i don't know when to use which formula.. i think one setis for connected batteries and one isn't. any help would beappreciated. cheers

Explanation / Answer

Charge stored on the capacitor Q = CV                                                    = 20 * 10 ^ -3 F * 150 V                                                    = 3 C (B). energy stored  Uo = ( 1/ 2) C V ^ 2                                 = 225 J (C). If the battery is connected with capacitor thenenergy U = K Uo If the battery is disconnect with capacitor then energy U = U o / K where K = dielectric constant = 4.3

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