Question

The gap between the plates of a parallel plate capacitor of area $ A $ and distance between plates, is filled with a dielectric whose permittivity varies linearly from $ {\varepsilon _1} $ at one plate to $ {\varepsilon _2} $ at the other. The capacitance of the capacitor is:
A) $ \dfrac{{{\varepsilon _0}({\varepsilon _1} + {\varepsilon _2})A}}{d} $
B) $ \dfrac{{{\varepsilon _0}({\varepsilon _1} + {\varepsilon _2})A}}{{2d}} $
C) $ {\varepsilon _0}A/\left[ {d/n\left( {\dfrac{{{\varepsilon _2}}}{{{\varepsilon _1}}}} \right)} \right] $
D) $ {\varepsilon _0}({\varepsilon _1} - {\varepsilon _2})A/\left[ {d/n\left( {\dfrac{{{\varepsilon _2}}}{{{\varepsilon _1}}}} \right)} \right] $

Answer 1

Hint: Since the dielectric constant of the material between the capacitors varies, find the potential drop across the two plates of the capacitor by integrating the potential drop across a small layer of the dielectric. Then use the relation between charge and voltage between two capacitor plates to find the capacitance.

Formula used: $ dV = - Edx $ where $ dV $ is the potential drop across a region of width $ dx $ with an electric field $ E $ .

Complete step by step solution:
Since the permittivity of dielectric varies linearly from $ {\varepsilon _1} $ at one plate to $ {\varepsilon _2} $ at the other we can define the permittivity at a distance $ x $ from one plate as:
 $ k = \left( {\dfrac{{{\varepsilon _2} - {\varepsilon _1}}}{d}} \right)x + {\varepsilon _1} $ where $ d $ is the distance between the two plates.
Since there is a dielectric present between the plates, the electric field in the region will be $ {E_0}/k $ where $ {E_0} = \dfrac{\sigma }{{{\varepsilon _0}}} $ is the electric field between the two plates of the capacitor when there is no dielectric present.
We can calculate the potential difference between the two plates using the relation $ dV = - Edx $ as
 $ dV = \dfrac{{ - {E_0}}}{k}dx $
On integrating both sides with the plates as the limits, we can write
 $ \int\limits_0^V {dV} = \int\limits_0^d {\dfrac{{ - {E_0}}}{k}dx} $
On substituting the value of the permittivity, we can write
 $ \int\limits_0^V {dV} = \int\limits_0^d {\dfrac{\sigma }{{{\varepsilon _0}}}\dfrac{1}{{\left( {\dfrac{{{\varepsilon _1} - {\varepsilon _2}}}{d}} \right)x + {\varepsilon _1}}}dx} $
On evaluating the integral, we get
 $ V = \dfrac{{\sigma d}}{{{\varepsilon _0}({\varepsilon _1} - {\varepsilon _2})}}\ln \left( {\dfrac{{{\varepsilon _2}}}{{{\varepsilon _1}}}} \right) $
We can then calculate the capacitance of the capacitor using the relation $ Q = CV $ where $ Q = \sigma A $ is the charged stored in the capacitor as
 $ C = \dfrac{Q}{V} = \dfrac{{A\sigma }}{{\dfrac{{d\sigma }}{{{\varepsilon _0}({\varepsilon _1} - {\varepsilon _2})}}\ln \left( {\dfrac{{{\varepsilon _2}}}{{{\varepsilon _1}}}} \right)}} $
 $ \Rightarrow C = \dfrac{{{\varepsilon _0}({\varepsilon _1} - {\varepsilon _2})A}}{{d\ln \left( {\dfrac{{{\varepsilon _2}}}{{{\varepsilon _1}}}} \right)}} $
which corresponds to option (D).

Note:
Whenever the dielectric varies with the space between the dielectrics, we should remember to find the electric fields between the two plates for an arbitrarily chosen potential between the two plates. The choice of the arbitrarily chosen potential won’t matter since it gets cancelled out in the end as the capacitance of a capacitor only depends on the physical dimensions of the plates and the permittivity of the dielectric.