Question

A capacitor is formed by two square metal-plates of edge $a$, separated by a distance$d$. Dielectric of dielectric constants $K$ is filled in the gap as shown in the figure. The equivalent capacitance is

• A. $\frac{k{\epsilon }_0{a}^2\ln k}{d\left(k-1\right)}$
• B. $\frac{k{\epsilon }_0{a}^2\ln k}{d\left(k-2\right)}$
• C. $\frac{k{\epsilon }_0{a}^2\ln k}{2d\left(k-1\right)}$
• D. $\frac{2k{\epsilon }_0{a}^2\ln k}{d\left(k-1\right)}$

Answer 1

The correct option is A

$\frac{k{\epsilon }_0{a}^2\ln k}{d\left(k-1\right)}$

Step 1: Given data

Length of the metal edge is $a$ and separation is $d$

Step 2: Solving for total capacitance:

This configuration can be considered as capacitors in series of edge $dx$ each at distance $x$ from one end with dielectric material increasing from one end to another, having a capacitance of $d{C}_{net}$.

$\frac{1}{d{C}_{net}}=\frac{d_1}{{\epsilon }_0{A}_1}+\frac{d_2}{k{\epsilon }_0{A}_2}$

Where $d_1$ is the distance from the first end and $d_2$ from another $A_1$and $A_2$ are the corresponding area, ${\epsilon }_0$ is permeability in a vacuum and $k$ is dielectric constant

Assuming that the angle made by dielectric filling at one end is $\varphi$

$d_1=d-x\tan \varphi$ and $d_2=x\tan \varphi$

where $\tan \varphi =\frac{d}{a}$

The area of there capacitance is, $edge\times dx=adx$

$\frac{1}{d{C}_{net}}=\frac{d-x\tan \varphi }{{\epsilon }_{0}adx}+\frac{x\tan \varphi }{k{\epsilon }_{0}adx}$

$d{C}_{net}=\frac{{\epsilon }_{0}adx}{d-x\tan \varphi +\frac{x\tan \varphi }{k}}$

Integrating with respect to$x$ with limits$0toa$

$d{C}_{net}={\epsilon }_{0}ka{\int }_0^a\frac{dx}{kd-xk\tan \varphi +x\tan \varphi }$

$\Rightarrow C_{net}=\frac{{\epsilon }_{0}ka}{\left(1-k\right)\tan \varphi }\left(0-\log k\right)$

$\Rightarrow C_{net}=\frac{{\epsilon }_{0}ka}{\left(1-k\right){\displaystyle \frac{d}{a}}}\left(0-\log k\right)$

$\Rightarrow C_{net}=\frac{{\epsilon }_{0}k{a}^2}{\left(k-1\right){\displaystyle d}}\log k$

Hence, option (A) is correct.