Question 106: Electrostats & Capacitance

Question

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, $$\delta  = {d \over N}$$. The dielectric constant of the mth layer is $${K_m} = K\left( {1 + {m \over N}} \right)$$. For a very large N(>103), the capacitance C is $$\alpha \left( {{{K{\varepsilon _0}A} \over {d\ln 2}}} \right)$$The value of $$\alpha $$ will be ..................[$$ \in $$0 is the permittivity of free space.]

Accepted Answer

Parallel plate capacitor,
$${x \over m} = {d \over N}$$
$$d\left( {{1 \over C}} \right) = {{dx} \over {{K_m}{\varepsilon _0}A}} = {{dx} \over {K{\varepsilon _0}A\left( {1 + {m \over N}} \right)}} = {{dx} \over {K{\varepsilon _0}A\left( {1 + {x \over d}} \right)}}$$
Integration on both sides, we get
$${1 \over {{C_{eq}}}} = \int {d\left( {{1 \over C}} \right) = \int_0^D {{{d\,dx} \over {K{\varepsilon _0}A(d + x)}}} } $$
$${1 \over {{C_{eq}}}} = \int {{d \over {K{\varepsilon _0}A}}\ln 2}  \Rightarrow {C_{eq}} = {{K{\varepsilon _0}A} \over {d\,\ln 2}}$$
Therefore, $$\alpha $$ = 1.

Electrostats & Capacitance

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