IIT JEE , AIEEE Forum - Physics , Electricity

perin

a parallel plate capacitor of capacitance C (without dielectrics ) is filled by dielectric slabs as shown in figure. then the new capacitance of the capacitor is:
a 3.9C
b 4C
c 2.4C
d 3C

karthik2007

Easy!!!

Hint:

The two small capacitors are in parallel (at the bottom), which in turn is in series with the one at the top.

perin

@karthik

wat's yur answer

aggmegha90

my ans is 4C. is it right??

perin

it's wrong........!!!!!

elastiboysai

ans is 3c
as karthik said
2c and 4c are in parallel so c eff for dem is 6c
now d 2 6c s are in series,
6c*6c/12c
3c

rtiit

@elastiboysai

how will the capacity of upper half be 6C??

it should be 12C........

plz check once again

perin

@elastiboysai yur answer is also wrong....!!

this question came in my iit pattern test and i marked 4C as answer and in fact i was thinking why did anyone put this soooo very easy question in the IIT PATTERN TEST but was shocked to see the answer...
it's 3.9 C

in the solution booklet the solution is given but i can't understand it they have written that the two small capaitors are not at equal potential difference so they can't be taken as parallel.....

can sumbdy explain this....pleeeeez

ashish_banga

are the lower two dielectrics have equal area

perin

yes i think they have equal area

bharat.singh

3.9C is the right ans.
as the lower two dielectrics will provide two way of induction of charges in the capacitor
one from k=2 and one from k=4
and then they will be sepratelty induced to k=6 with opposite charge

now try it and tell me

perin

bharat plz explain again ... i don't get it

karthik2007

You guys have probably mixed up the areas and the widths of the plates. Please check your calculations

v.sabarish

There is no reason why the two small capacitors must be at the same potential difference.......
while solving such problems, we assume a thin metal plate between the dielectrics and take them as combinations of capacitors..........
a metal plate implies same potential at the interface which in turn means the potential difference across the two capacitors is the same.....
but there is no plate, so no question of taking equal potentials or potential differences......
as the solution goes, divide the larger capacitor into two halves... each of capacitance 6 C.......
then 2,6 in series give 2*6/2+6=12/8 and 4,6 give 4*6/4+6=24/10
these in parallel...
so, they add up to give 39/10or 3.9 C

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