A parallel plate capacitor has plates of area 'A' separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x)=K(1+αx) where 'x' is the distance measured from one of the plates. If (αd)<<1, the total capacitance of the system is best by the expression:
A
dAϵ0K(1+(2αd)2)
B
dAϵ0K(1+2α2d2)
C
dAϵ0K(1+2αd)
D
dAϵ0(1+αd)
Hard
JEE Mains
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Updated on : 2022-09-05
Solution
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Correct option is C)
Refer image 1.
Let k(x)=k(1+αx)
be the dielectric constant
The capacitor can be considered to be a series combination of a large number of capacitors of thickness Δx.
Refer image 2.
Ceq1=∑k(x)Aϵ0Δx=∫0dk(x)Aϵ0dx
where Δx→0
Ceq1=∫0dk(1+αx)Aϵ0dx=KAϵ01∫0d1+αxdx
=KAϵ01loge(11+αd)×α1
=KαAϵ01loge(1+αd)
Now, log(1+x)=x−2x2+3x3−−−−−−−
∴loge(1+αd)≃αd−2α2d2
∴Ceq1=kαAϵe1loge(1+αd)
=kαAϵ01[αd−2α2d2+−−−−−−−]
≃kαAϵ0αd[1−2αd]≃KAϵ0d[1−2αd]
Ceq=d[1−2αd]KAϵ0=dKAϵ0[1−2αd]−1
Now, (1−x)−1≃1+x[∵x<<1]
∴Ceq=dKAϵ0[1+2αd]
option (C) is correct.
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