1i want the matches..
woh sab toh mujhe bhi pata hai [3]
21. A parallel plate capacitor with air gap between its plates is charged using a battery and remains connected to the battery. Match the column I with column II.
Column I
Column II
(a) Electric field intensity between the plates reduces when (p) separation between the plates is increased
(b) Electric energy stored in the capacitor increase when (q) separation between the plates is reduced
(c) electric energy density increases when
(r) dielectric of dielectric constant K is filled between the plates
(d) Charge on the capacitor decreases when (s) the separation between the plates is increased by the 3/2 times the initial value and the space between the plates is then completely filled with a dielectric of dielectric constant K
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10 Answers
39when a dielestric is intoduced, the capacitance is increased by a factor of k
and the electric field, potentiall and potential energy are decreased by a factor of k.
24a q
b q,r
c p,s
and i cant understand last line..is it in continuation with option d?
1am afraid! none of my ans matched wid urs.!
i did
a- r,s
b- p,s
c- q
d- ka toh kuchh nahi mila :(
24ONE OF US HAS TO BE WRONG[6]
electric field =potential diff * distance
so it reduces when separation is reduced
electric energy stored=1/2 C V2
so it increses whne capacitane inc. i.e when separation decrases or area inc or dielectric insertes
62@ eureka.. think of gauss's law.
derivation of electric field for a parallel plate capacitor.
1a ---------------> p
b --------------->q,r
c ---------------->q,r
d ---------------->p,to consider s option, value of K is required