When a capacitor is filled with dielectric slab and the capacitor is also connected to a battery,then due to dielectric slab,the potential difference across capacitor decreases but the decrease in potential difference is taken into account by the battery .Battery tries to maintain the same potential difference .But if the potential difference is same across the capacitor as the previous one then why should capacitance increase ? I mean to say that we usually say that insertion of dielectric makes potential difference across the capacitor less and since Q=C/V, charge increases ,thus capacitance increases.Please reply,I am very confused.
what i think is
that as the dilectric is inserted the capacitance increases by an amount 'K' but since battery has to maintain same pot. diff therefore a charge kQ is supplied thus charge/capacitance reamins same.
i hope i cleared ur doubt
Oh c'mon you cant be seriously confused on such a silly thingIf you are unable to understand something then why dont you go from basics da.Just draw 2 parallel plates put a dielectric slab in between and write the gauss law equation for feild between the 2 plates find the potential diff between the 2 plates using F.S as force is a constant.Now equate this expression to the p.d across your battery & you would get a relation between the charge on the plate, the p.d across ur battery. Now should be able 2 know why charge increases .(dont forget 2 include k in your field eqn) One more thing stupid girl(since you consider yourself to be stupid) What you have written in the last 3 lines of your question is absolutey stupid , Just give me the name of the fellow who talked like that.
Its not at all that way girl just write the equations having a slab and removing the slab and having the same battery you observe that the former has 'k' times the charge of the latter and k>1. And that optics question are you serious you didnt understand the question? God i thought that you were a real clever girl....
Its not at all that way girl just write the equations having a slab and removing the slab and having the same battery you observe that the former has 'k' times the charge of the latter and k>1. And that optics question are you serious you didnt understand the question? God i thought that you were a real clever girl....
Manoj, actually question said that ray intersects the diameter at the point of emergence ,I was not able to imagine the situation.