A parallel plate capacitor is made of two square plates of side $$a$$, separated by a distance $$d$$ ($$d \ll a$$). The lower triangular portion filled with a dielectric of dielectric constant $$K$$, as shown in the figure. Capacitance of this capacitor is:

We have two large square plates of side $$a$$ separated by a small distance $$d$$, with $$d \ll a$$ so that edge effects can be ignored and the electric field between the plates is practically uniform in the vertical direction. The space between the plates is partly air and partly filled with a dielectric of constant $$K$$. The interface between the dielectric and the air is a straight line joining the bottom left corner of the plates to the top right corner, so that at the left edge there is no dielectric while at the right edge the entire gap is filled with dielectric. In the cross-section (an $$x\!-\!y$$ plane) the dielectric therefore occupies the lower triangular region, and the air occupies the upper triangular region.

To find the total capacitance we slice the plates into thin vertical strips, each of width $$dx$$, parallel to the interface. Every strip behaves as an independent capacitor because all strips have the same potential difference $$V$$ across the plates. Hence the capacitances of the strips are in parallel and will add algebraically.

For a strip situated at a horizontal coordinate $$x$$ (measured from the left edge, $$0 \le x \le a$$), the diagonal interface meets the strip at a height that is proportional to $$x$$. Since the interface goes from the bottom left corner $$(x=0,y=0)$$ to the top right corner $$(x=a,y=d)$$, the thickness of the dielectric layer inside this strip is

$$t_2=\frac{x}{a}\,d,$$

and the remaining thickness is air, namely

$$t_1=d-t_2=d-\frac{x}{a}d=d\Bigl(1-\frac{x}{a}\Bigr).$$

Thus, along the field direction, the air layer and the dielectric layer are connected in series. For a series combination of two capacitors of plate area $$A$$ and thicknesses $$t_1$$ and $$t_2$$ with dielectric constants $$\varepsilon_1$$ and $$\varepsilon_2$$, the formula is first stated:

$$\frac{1}{C_{\text{series}}}=\frac{t_1}{\varepsilon_1 A}+\frac{t_2}{\varepsilon_2 A}.$$

Here $$\varepsilon_1=\varepsilon_0$$ (air), $$\varepsilon_2=K\varepsilon_0$$ (dielectric), and the area of the strip is

$$A= (\text{depth})\times (\text{width}) = a\;dx.$$

Substituting these into the formula we get

$$\frac{1}{dC(x)}=\frac{t_1}{\varepsilon_0 A}+\frac{t_2}{K\varepsilon_0 A}

=\frac{d\bigl(1-\dfrac{x}{a}\bigr)}{\varepsilon_0 a\,dx}

+\frac{d\dfrac{x}{a}}{K\varepsilon_0 a\,dx}.$$

Combining the two fractions in the numerator:

$$\frac{1}{dC(x)}=\frac{d}{\varepsilon_0 a\,dx}

\Bigl[\,1-\frac{x}{a}+\frac{x}{aK}\Bigr]

=\frac{d}{\varepsilon_0 a\,dx}

\Bigl[1-\frac{x}{a}\Bigl(1-\frac{1}{K}\Bigr)\Bigr].$$

Hence the capacitance of the strip is

$$dC(x)=\frac{\varepsilon_0 a\,dx}

{d\Bigl[1-\dfrac{x}{a}\Bigl(1-\dfrac{1}{K}\Bigr)\Bigr]}.$$

For convenience define

$$\alpha=1-\frac{1}{K},$$

so that

$$dC(x)=\frac{\varepsilon_0 a\,dx}{d\bigl[1-\dfrac{\alpha x}{a}\bigr]}.$$

All these strip capacitances are in parallel, so the total capacitance is obtained by integrating $$dC(x)$$ from $$x=0$$ to $$x=a$$:

$$C=\int_{0}^{a} dC(x)=\frac{\varepsilon_0 a}{d}\int_{0}^{a}

\frac{dx}{1-\dfrac{\alpha x}{a}}.$$

We now carry out the integral. First rewrite the denominator:

$$1-\frac{\alpha x}{a}=1-\alpha\left(\frac{x}{a}\right).$$

Let us substitute

$$u=1-\alpha\left(\frac{x}{a}\right)\quad\Longrightarrow\quad

du=-\frac{\alpha}{a}\,dx\quad\Longrightarrow\quad

dx=-\frac{a}{\alpha}\,du.$$

When $$x=0$$, $$u=1$$; when $$x=a$$, $$u=1-\alpha.$$ Therefore

$$

\begin{aligned}

C & =\frac{\varepsilon_0 a}{d}\int_{x=0}^{x=a}

\frac{dx}{1-\dfrac{\alpha x}{a}}

=\frac{\varepsilon_0 a}{d}\int_{u=1}^{u=1-\alpha}

\frac{-\dfrac{a}{\alpha}\,du}{u} \\

& =\frac{\varepsilon_0 a}{d}\cdot\frac{a}{\alpha}

\int_{u=1-\alpha}^{u=1}\frac{du}{u} \\

& =\frac{\varepsilon_0 a^2}{d\alpha}

\bigl[\ln u\bigr]_{u=1-\alpha}^{u=1}.

\end{aligned}

$$

Evaluating the limits:

$$

\begin{aligned}

\bigl[\ln u\bigr]_{1-\alpha}^{1}

&=\ln(1)-\ln(1-\alpha)\\

&=0-\ln(1-\alpha)\\

&=-\ln(1-\alpha).

\end{aligned}

$$

So

$$C=\frac{\varepsilon_0 a^2}{d\alpha}\,\bigl[-\ln(1-\alpha)\bigr]

=\frac{\varepsilon_0 a^2}{d\alpha}\,\ln\!\Bigl(\frac{1}{1-\alpha}\Bigr).$$

Recall that $$\alpha=1-\dfrac{1}{K} \; \Rightarrow \; 1-\alpha=\dfrac{1}{K}.$$ Substituting this back in gives

$$

C=\frac{\varepsilon_0 a^2}{d\alpha}\,\ln(K)

=\frac{\varepsilon_0 a^2}{d}\;

\frac{1}{1-\dfrac{1}{K}}\,\ln K.

$$

Simplifying the fraction:

$$

\frac{1}{1-\dfrac{1}{K}}

=\frac{1}{\dfrac{K-1}{K}}

=\frac{K}{K-1}.

$$

Therefore

$$

C=\frac{\varepsilon_0 a^2}{d}\,

\frac{K}{K-1}\,\ln K

=\frac{K\varepsilon_0 a^2}{d\,(K-1)}\,\ln K.

$$

This matches Option 2 in the given list.

Hence, the correct answer is Option 2.