Work done on capacitors and more

AI Thread Summary
The discussion revolves around a physics problem involving a capacitor with a conducting plate inserted between its plates. Participants analyze how the capacitance, charge, and energy change when the plate is added and subsequently removed. Key calculations include determining the new capacitance as 3C/2, the increase in charge as CE/2, and the increase in energy as CE²/4. The work done by the battery during the insertion is calculated as CE²/2, while the work done by the force during insertion and removal is noted as negative changes in energy. A question remains regarding whether the voltage changes when the conductor is removed.
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Homework Statement



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?

Homework Equations



C=Q/V
U=q²/2C
C=\epsilon0A/d

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!
 
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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
 
Ohh, I got it!
(1) Initial capacitance: 0A/d
When we introduce the conductor, each of the "two capacitors" will have capacitance = 3\epsilon0A/d. Since I can assume the "two capacitors" are arranged in series, Final capacitance = Capacitance/2. So, final capacitance will be 3\epsilon0A/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! ^^
 
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