Question

A parallel-plate capacitor has capacitance 5.20 μF. The capacitor was origionaly connected to a 1.50 V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? Note: When disconnected, the charge on the capacitor must remain the same as when disconnected.

A parallel-plate capacitor has capacitance 5.20 μF. The capacitor was origionaly connected to a 1.50 V battery? (c) The battery is subsequently reattached to the capacitor, but the plate separation remains doubled as in part (b). How much energy is stored?

Answer 1

C= 5.20 μF.

V= 1.50

so Q= C*V = 5.20*1.50 = 7.8μC

now distance between the charged plates doubled d' = 2*d  

only capacitance changes because C= Aε/d so

because d'= 2*d so new C' = C/2 = 5.20/2 = 2.6 μF

so energy = Q^2 /2*C' =  (7.8μC )^2/ ( 2*2.6 μF) = 11.7 μJ = 11.6*10^-6 J = answer

(c)

energy = 1/2*C' *V^2  

C' = 2.6 μF

V= 1.5 V

so energy =1/2*2.6*1.5^2 =2.925 *10^-6 J = 2.925 μJ answer

Good Luck!!

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