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Q. Find the total potential energy gained by a capacitor which is connected to a battery of emf V as a dielectric slab of dielectric constant K is allowed to enter slowly into the capacitor. Assume it is a parallel plate capacitor with the distance between the plates to be d. The dimensions of the plate are l x b . Also calculate the energy lost by the battery.
P.S I need a lot of help in this one. Tried it a lot.
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3 Answers
3when the slab is put in fully ... the cap. increses to kC ... but the pot diff remains same ....
so the q on the cap changes .........
q1/C = q2/kC .................
q1 = q2/k .............
q2 = k q1 .............
change in energy of system ..........
1/2(C(k-1))V^2 ..........................(1)
amt of energy used by batery used to mobilise excess charge ......
(k-1)(CV)(V) ..................(2)
AMT OF ENERGY LOST BY BATERRY .........
1/2(CV^2(K-1)) ........................ [[[ (2)-(1) ]]]
the energy lost???????
this means that the work done by the battery is not fully used to increse pot energy of the system ........... so the extra engy goes out as heat :((((( ..................
hope u understood asish////////////