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, $K\left(x\right)=K_o+\lambda x(\lambda =a$ constant) The capacitance C, of this capacitor, would be related to its 'vacuum' capacitance $C_o$ as per the relation :

• A. $C=\frac{\lambda d}{ln(1+k_o\lambda d)}{C}_o$
• B. $C=\frac{\lambda }{d.ln(1+k_o\lambda d)}{C}_o$
• C. $C=\frac{\lambda d}{ln(1+\lambda d/K_o)}{C}_o$
• D. $C=\frac{\lambda }{d.ln(1+k_o/\lambda d)}{C}_o$

Answer 1

Answer: (C)

$C=\frac{\lambda d}{ln(1+\lambda d/K_o)}{C}_o$

Capacitance of the vacuum capacitor $C_o=\frac{A{ϵ}_o}{d}$
where $A$ is the area of the plates of capacitor.
Dielectric constant of the medium at a distance $x$ from first plate is given by $K\left(x\right)=K_o+\lambda x$
Capacitance of capacitor of thickness $dx$, $dC=\frac{K\left(x\right)A{ϵ}_o}{dx}=\frac{A{ϵ}_o(K_o+\lambda x)}{dx}$
Total capacitance $\frac{1}{C}={\int }_o^d\frac{dx}{A{ϵ}_o(K_o+\lambda x)}$
Or $\frac{1}{C}=\frac{1}{A{ϵ}_o}\times \frac{\ln (K_o+\lambda x)}{\lambda }{|}_o^d$
Or $\frac{1}{C}=\frac{1}{A{ϵ}_o\lambda }\ln (1+\lambda d/K_o)$
Or $C=\frac{A{ϵ}_o\lambda }{\ln (1+\lambda d/K_o)}$
$⟹$ $C=\frac{\lambda d}{\ln (1+\lambda d/K_o)}{C}_o$