IIT JEE , AIEEE Forum - Physics , Electricity

neeraj_agarwal_1990

B/w the plates of a parallel plate capacitor of capacitance C, 2 parallel plates , of same material and area same as plate of the original capacitor are placed. If the thickness of these plates is equal to 1/5 th of the distance b/w the plates of the original capacitor , then the capacity of the new capacitor is???

ans: 3/5C

rooney

When 2 plates of thickness d/5 are inserted, there are 3 capacitors formed.
With distance between each capacitor being d/5.
Let C' = Capacitor of 1 capacitor = kE₀A/(d/5) = 5C
These 3 capacitors are in series. So, net capacitance is C'/3 = 5C/3

Check your answer please. It should be 5C/3 and not 3C/5

neeraj_agarwal_1990

dats y i posted this quesion...
me too gettin' 5/3C!

rooney

Then the answer given is wrong.

rooney

EDIT: Posted in wrong thread. was posting somewhere else.

spideyunlimited

C = E_oA / d

C' = E_oA / (d/5)

:.

C'_net = C' / 3 .. series connection

= 5 E_oA / 3d

= 5C/3

joyfrancis

Hey!!, thickness of the plates and the distance between the plates are different things. If thickness of the plates is only given how are you calculating the capacitance??
I'm not getting this.

rooney

initial distance between two plates is d. Now, when you insert 2 plates of thickness d/5 , this 2xd/5 thickness is taken from the initial distance b/ew the plates d only. So, remaining space = d - 2xd/5 = 3d/5. There are there capacitors formed by the 4 plates. Distance between each pair of plates is considered to be constant and given by (3d/5)/3=d/5.

spideyunlimited

@joy.. yaar out of total distance d which is available, the thickness of the plates occupies some length so distance between plates finally is less.

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