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Question

(C) The dielectric constant is given by $K(x) = K_0 + \lambda x$.Consider a small element of thickness $dx$ at distance $x$. This acts as a capacitor

Vedclass · Accepted Answer

$C = \frac{\lambda d}{\ln(1 + \lambda d / K_0)} C_0$

Vedclass · Answer

(C) The dielectric constant is given by $K(x) = K_0 + \lambda x$.Consider a small element of thickness $dx$ at distance $x$. This acts as a capacitor with capacitance $dC = \frac{\epsilon_0 K(x) A}{dx}$.Since these elements are in series,the total capacitance $C$ is given by $\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{\epsilon_0 A (K_0 + \lambda x)}$.Evaluating the integral: $\frac{1}{C} = \frac{1}{\epsilon_0 A} \int_0^d \frac{dx}{K_0 + \lambda x} = \frac{1}{\epsilon_0 A \lambda} [\ln(K_0 + \lambda x)]_0^d$.$\frac{1}{C} = \frac{1}{\epsilon_0 A \lambda} \ln \left( \frac{K_0 + \lambda d}{K_0} \right) = \frac{1}{\epsilon_0 A \lambda} \ln \left( 1 + \frac{\lambda d}{K_0} \right)$.Since vacuum capacitance $C_0 = \frac{\epsilon_0 A}{d}$,we have $\epsilon_0 A = C_0 d$.Substituting this: $\frac{1}{C} = \frac{1}{C_0 d \lambda} \ln \left( 1 + \frac{\lambda d}{K_0} \right)$.Therefore,$C = \frac{\lambda d}{\ln(1 + \lambda d / K_0)} C_0$.

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