Question

The space between the plates of a parallel plate capacitor is filled with a ‘dielectric’ whose ‘dielectric constant’ varies with distance as per the relation, $K(x) = {K_0} + \lambda x$($\lambda$= a constant). The capacitance C, of this capacitor, would be related to its ‘vacuum’ capacitance ${C_0}$ as the relation :
(A) $C = \dfrac{{\lambda d}}{{\ln (1 + {K_0}\lambda d)}}{C_0}$
(B) $C = \dfrac{\lambda }{{d.\ln (1 + {K_0}\lambda d)}}{C_0}$
(C) $C = \dfrac{{\lambda d}}{{\ln (1 + \lambda d/{K_0})}}{C_0}$
(D) $C = \dfrac{\lambda }{{d.\ln (1 + {K_0}/\lambda d)}}{C_0}$

Answer 1

Dielectric constant, also known as relative permittivity or specific inductive capacity, is the property of an insulating material or dielectric which is equal to the ratio of the capacitance of a capacitor filled with the given dielectric material to the capacitance of an identical capacitor present in a vacuum without the dielectric material. In this question, the dielectric constant is varying with the distance so we first need to divide the material into parallel layers of width $dx$ having capacitance $dC$. The total capacitance will be the series equivalent capacitance of these layers.

Formulae used:
$C = \dfrac{{{K_0}A}}{d}$,
$\dfrac{1}{{{C_{eq}}}} = \int {\dfrac{1}{{dC}}}$

Complete step by step solution:
Capacitance of capacitor with dielectric material = ${C_0} = \dfrac{{{\varepsilon _0}A}}{d}$
Where, A = area of capacitor, and d = distance between capacitor plates.
Capacitor of capacitor layer of thickness ‘$dx$’ will be, (substituting the value for dielectric constant relation with the distance that is, $K(x) = {K_0} + \lambda x$ where $\lambda$ =constant),
$dc = \dfrac{{{\varepsilon _0}({K_0} + \lambda x)A}}{{dx}}$
Total capacitance will be,
$\dfrac{1}{C} = \int {\dfrac{1}{{dC}}}$
Putting the value of ‘$dC$’ and taking limits from ‘$0$’ to ‘$d$’.
$\dfrac{1}{C} = \int\limits_0^d {\dfrac{{dx}}{{{\varepsilon _0}({K_0} + \lambda x)A}}}$
$\dfrac{1}{C} = \dfrac{1}{{{\varepsilon _0}A}}\int\limits_0^d {\dfrac{{dx}}{{({K_0} + \lambda x)}}}$
On Integrating,
$\dfrac{1}{C} = \dfrac{1}{{A{\varepsilon _0}\lambda }} \times [\ln ({K_0} + \lambda x)]_0^d$
$\dfrac{1}{C} = \dfrac{1}{{A{\varepsilon _0}\lambda }} \times [\ln ({K_0} + \lambda d) - \ln ({K_0})]$
$\dfrac{1}{C} = \dfrac{1}{{A{\varepsilon _0}\lambda }} \times [\ln (1 + \dfrac{{\lambda d}}{{{K_0}}})$
$C = \dfrac{{A{\varepsilon _0}\lambda }}{{\ln (1 + \dfrac{{\lambda d}}{{{K_0}}})}}$
Multiplying by $d$ both numerator and denominator.
$C = \dfrac{{A{\varepsilon _0}\lambda d}}{{d\ln (1 + \dfrac{{\lambda d}}{{{K_0}}})}}$
$C = \dfrac{{\lambda d}}{{\ln (1 + \dfrac{{\lambda d}}{{{K_0}}})}}{C_0}$

Hence, the correct answer is option (C).

Note: The different layers of dielectric material are placed parallel to each other but in the circuit, they will appear to be in series to each other. Hence we had to use a formula for capacitance in series and not that of capacitance in parallel. When a dielectric slab is inserted in the capacitor, the capacitor increases by a factor K, where K is the dielectric constant.