Question
Hey everyone,
I'm having difficulty understanding this question:
Background:
There are two identical capacitors with capacitance C. Thefirst capacitor is charged with charge Q and has a potentialdifference of V across its plates. The second capacitor isuncharged. The second capacitor is then connected to thefirst - the left plate to the left plate and the right plate to theright plate. As a result, the second capacitor becomescharged.
True or False?
Because energy is conserved, the energy in the two capacitors(after they are connected) equals the energy stored in the firstcapacitor (before they are connected).
Please explain your reasoning, I would very much like to understandthe answer. Thank you so much!
Explanation / Answer
Common potential after connected V ' = [ C V + C (0 ) ] / ( C + C ) = V / 2 Since the two capacitors are connected to parallel. Energy stored in first capacitor( before they are connected) E = ( 1/ 2) C V 2 Energy stored in second capacitor( before they areconnected ) E = 0 Total energy before connected = ( 1/ 2) C V2 Energy stored in second capacitor( before they areconnected ) E = 0 Total energy before connected = ( 1/ 2) C V2 Energy stored in first capacitor( after they areconnected ) E ' = ( 1/ 2) C ( V/2) 2 = ( 1/ 4 ) ( 1/ 2) C V 2 Energy stored in second capacitor( after theyare connected ) E " = ( 1/ 2) C ( V/2) 2 = ( 1/ 4 ) ( 1/ 2) C V 2 Total energy after connected = ( 1/ 4 ) ( 1/ 2) C V2 + ( 1/ 4 ) ( 1/ 2) C V 2 = ( 1/ 2 ) ( 1/ 2) C V 2 = ( 1/ 4 ) C V 2 So, energy not conserved = ( 1/ 4 ) ( 1/ 2) C V 2 Energy stored in second capacitor( after theyare connected ) E " = ( 1/ 2) C ( V/2) 2 = ( 1/ 4 ) ( 1/ 2) C V 2 Total energy after connected = ( 1/ 4 ) ( 1/ 2) C V2 + ( 1/ 4 ) ( 1/ 2) C V 2 = ( 1/ 2 ) ( 1/ 2) C V 2 = ( 1/ 4 ) C V 2 So, energy not conserved So, energy not conserved