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QJEEAdv19Ph608d: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph608d Solution: A parallel plate capacitor of capacitance C has spacing d…

A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness, δ=dN\delta = {d \over N} . The dielectric constant of the mth layer is Km=K(1+mN){K_m} = K\left( {1 + {m \over N}} \right) . For a very large N(>103), the capacitance C is α(Kε0Adln⁡2)\alpha \left( {{{K{\varepsilon _0}A} \over {d\ln 2}}} \right)

The value of α\alpha will be ..................

[ ∈\in 0 is the permittivity of free space.]

Step-by-Step Verified Solution

Parallel plate capacitor,

xm=dN{x \over m} = {d \over N}

d(1C)=dxKmε0A=dxKε0A(1+mN)=dxKε0A(1+xd)d\left( {{1 \over C}} \right) = {{dx} \over {{K_m}{\varepsilon _0}A}} = {{dx} \over {K{\varepsilon _0}A\left( {1 + {m \over N}} \right)}} = {{dx} \over {K{\varepsilon _0}A\left( {1 + {x \over d}} \right)}}

Integration on both sides, we get

1Ceq=∫d(1C)=∫0Dd dxKε0A(d+x){1 \over {{C_{eq}}}} = \int {d\left( {{1 \over C}} \right) = \int_0^D {{{d\,dx} \over {K{\varepsilon _0}A(d + x)}}} }

1Ceq=∫dKε0Aln⁡2⇒Ceq=Kε0Ad ln⁡2{1 \over {{C_{eq}}}} = \int {{d \over {K{\varepsilon _0}A}}\ln 2} \Rightarrow {C_{eq}} = {{K{\varepsilon _0}A} \over {d\,\ln 2}}

Therefore, α\alpha = 1.

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