Capacitors 3

when 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////////////

put C as e(l*b)/d .... in the abv expression.....