IIT JEE , AIEEE Forum - Physics , Electricity

akhil_o

A parallel plate capacitor is made of square plates side length a

A parabolic slab of k=2 is inserted

equation of parabola is

x*x=ay and the vertex touches the plate in the iddle

Then:

1)This arrangement is equivalent to two capacitors CEOA and EDBO in parallel

2)This arrangement is equivalent to two capacitors CEOA and EDBO in series

3)Capacitance is 2

₀a

4)Capacitance is

₀a

akhil_o

Anybody?
if u dont hav time just tell me the concept and her to start off....no need to post the soln!

nandithac

pls show the terminals of the capacitor so that i can say if it is in parallel or in series

akhil_o

current flows across from plate CED to AOB

aditya_arora04

Answer is just option 1.

saurabh_reincarnated

as current flows frm CED to AOB the portions will be in parallel as they have same potential diff across them

now to fnd capacitace ....

use simple methud ...

take some dx elemnt at x distance...

(now the parabola must be vertical for it to be a function..bcos we will use integration of a function .....as shown in fig.)

now capacitance due to that dx element is due to two thngs in series that is

the shaded and unshaded part.. calculate capacitance due to dx element in terms of dx, y and also a... remeber they r in series so use that form......(1)

(connect all sections in series here becos pot diff . across them are not same and follow series rule)

now similarly all dx element ud follow that rule for cap.... and all dx elements are connected in parallel as they hve same pot diff. across them..

so C1+C2+C3....= net C or total capacitance.....

thus if we integrate it we will get eff cap (rather thn adding integrating a function is easy)......

use expression found in ...(1) to integrate.. and use that y= ax^2..and int. frm

x= -1/2 to +1/2.... this x is got by putting y=a/4 in eq... y=ax^2...

hence u will fnd effectve capacitance

hope u got it.......

Cheers!!!!!!

astrojith

Hey, can anyone post a cleaner figure. That thing is totally screwed.

saurabh_reincarnated

the figure is absolutely fine..... and clear..

it also describes all u need for the prob...

its just parallel plates of cap. in CED and AOB

and the dielectric in inserted as shaded portion as shown very clearly.......

nandithac

ANS IS OPTION 1

karthik2789

see i tell u the fundamental,
option 1 is correct and to calculate the capacitance , we use definite integral

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