wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A fully charged parallel plate capacitor is connected across an uncharged identical capacitor . Show that the energy stored in the combination is less than stored initially in the single capacitor

Open in App
Solution

Initially we consider a charge capacitor; then its charge would be
Q=cv
U1=12CV2 __(I)
Now the uncharged capacitor is connected to charged capacitor
Then charge flows from 1st capacitor to other capacitor unless both capacitor attain common potential
Q1=CV1 and Q2=CV2
Applying conservation of charge
Q=Q1+Q2CV=CV1+CV2
V=V1+V2V=2V1V1=V2
Total energy stored U2=12CV21+12CV22
U2=14CV2 __(II)
eq (II) < eq (I)
So U2<U1

1312059_1067155_ans_8b04908d1de546bf8a04bad13630e7a8.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Placement of Dielectrics in Parallel Plate Capacitor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon