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Well in this section i will post some lengthy but comprehensive and easy to tackle qs for others and i will give one week for soln.
This is the first one .
in the given figure the capacitor has a dielectric and attached to a resistor and cell. When the circuit is closed the current flows to the dielectric and it gets charged. Now the new thing is that the dielectric isn't a perfect insulator but has a finite resistance W while the dielectric constant is K for the same.
The capacitor is charged and also discharged. Assuming that rate of charging is more than rate of discharging. Find :
1 : charge on the plate as a function of time if w is tending to infinity.........3 marks
2 : charge as a function of time if w is a real finite....................5 marks
3 : verify your result of qs 1 by using the result of qs 2............2 marks
{ guidlines : if there is some error in qs nudge me. Please post all your answer together.}
date : 17th june, 2009
due time : 1 week
Comments (6)
WELL VIGNESH first of all one advice as a friend. Always try a qs first by the FUNDAMENTAL METHOD. Here I think you have made a fallacy but I will keep my words and post it after 4days as I had said.
The qs is very fundamental and easy. Not at all unncesary tough and I believe you will soon have qs on " leakage current " in future IIT JEE. We have qs like this in I.E. Irodov - I was inspired to make this qs from there.
keep trying!!!!!!!
as dieletric has sum finite resistanace so we have to consider drop in potential due to that resistance W also
applying kirchoff's rule across the loop
we get
E - i*W - i*R - q/c = 0
rearranging we get
= 
at t=0 considering q=0 becoz capacitor is uncharged initially
soving we get
hahahaha Now I CAME TO KNOW WHAT BLUNDER VIGNESH AND TARAN HAVE DONE. CURRENT IN DIELECTRIC AND CELL IS NOT SAME...........WHAT A MISTAKE ? THATS WHY I GAVE THE QS. i CURRENT IS CHARGING THE PLATE OTHER IS DISCHARGING.
sorry both of you.........but you are wrong
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yeah. gud quesn.. and gud work by u blade x.. u deserve a 100 cakes for starting this....
and now, to business...
hey, i dont know whether u're able to see the picture..
but here it is.. the resistance of the dielectric slab can be considered a resistance wire parallele to capacitor..
let R (eff) be the net resistance across capacitor,
so, 1/ R(eff) = 1 / R + 1/W
ques1 : applying kircoff's laws and deriving, i'm getting
that is.. q = EC { 1 - e^ [-(t/RC + t/WC)] }
where C = KEA/ l ( E = epsilon not )
ques 2 and 3 : put W = infinity
q = EC { 1 - e^(-t/RC) }