# Work done on capacitors and more

1. Jun 9, 2010

### Tntgsh

1. The problem statement, all variables and given/known data

Hello, guys! Here I am again asking for your help...This is question number 2 from http://www.studyjapan.go.jp/en/toj/pdf/007.pdf"
2) Consider the circuit shown in Fig. 5, consisting of a battery of voltage E, a switch S, and a parallel-plate capacitor with capacitance C. The capacitor consists of two parallel conducting plates of equal area A separated by a distance d. After the switch S is closed and the capacitor is fully charged, a conducting plate of thickness d/3 and area A is inserted slowly between the plates of the capacitor. The inserted conducting plate is kept parallel to the conducting plates of the capacitor. Select answers to the questions from (a) to (z) below, and write the symbol of the answer in the box.
(1) Find the capacitance of the capacitor after the conducting plate is inserted.
(2) How much is the increase in the charge stored in the capacitor caused by inserting the conducting plate?
(3) How much is the increase in the energy stored in the capacitor caused by inserting the conducting plate?
(4) How much work is done by the battery during the insertion of the conducting plate?
(5) How much work is done by the force applied to the conducting plate during its insertion?
Next, the switch S is opened, and the conducting plate is removed slowly. How much work is done by the force applied to the conducting plate to remove it?

2. Relevant equations

C=Q/V
U=q²/2C
C=$$\epsilon$$0A/d

3. The attempt at a solution

Actually, I haven't done much 'cause I don't know what to do. Any ideas?
Sorry about not posting anything at the "attempt" section, but I really don't know how to start...I tried, I swear! Actually, I tried a lot ><
Thanks for the help, guys!

Last edited by a moderator: Apr 25, 2017
2. Jun 9, 2010

### bharath423

when we insert a conductor in between due to field -ve charge comes on one side and +ve charge on the other and since its a conductor Electric field inside is zero so charge comes on to surface,so its just like two new capacitors with distance between palates is d/3 each..
potential drops as E/2,E/2 to each then u can find the charge on plates with new capacitance values for each.
so the we get the additional charge the battery should provide,so the work done by battery is additional charge*E...
to find work done by u find the change i potential nergy before and after placing conductor

3. Jun 12, 2010

### Tntgsh

Ohh, I got it!
(1) Initial capacitance: 0A/d
When we introduce the conductor, each of the "two capacitors" will have capacitance = 3$$\epsilon$$0A/d. Since I can assume the "two capacitors" are arranged in series, Final capacitance = Capacitance/2. So, final capacitance will be 3$$\epsilon$$0A/2d=3C/2.
(2)Initial charge = CE
Final charge = 3CE/2
Increase in charge = 3CE/2-CE=CE/2
(3)Initial energy = CE²/2
Final energy = 3CE²/4
Increase in energy = 3CE²/4-CE²/2=CE²/4
(4)bharath423 said that the work done by the battery will be Additional charge*E=CE²/2
(5)The work done by the force will be =-Increase in energy=-CE²/4
(6)The work done by the force will be, once again, =-Increase in energy.
Energy now=3CE²/4
But here's my doubt...when I remove the conductor, will the voltage E change?
Thanks a lot for the help! ^^