# A parallel plate capacitor is made of two square plates of side

Question:

A parallel plate capacitor is made of two square plates of side ' $a$ ', separated by a distance $d(d< 1. (1)$\frac{\mathrm{K} \in_{0} \mathrm{a}^{2}}{2 \mathrm{~d}(\mathrm{~K}+1)}$2. (2)$\frac{\mathrm{K} \in_{0} \mathrm{a}^{2}}{\mathrm{~d}(\mathrm{~K}-1)} \operatorname{In} \mathrm{K}$3. (3)$\frac{\mathrm{K} \in_{0} \mathrm{a}^{2}}{\mathrm{~d}}$In$\mathrm{K}$4. (4)$\frac{1}{2} \frac{K \in_{0} a^{2}}{d}$Correct Option: , 2 Solution: (2) From figure,$\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{a}} \Rightarrow \mathrm{y}=\frac{\mathrm{d}}{\mathrm{a}} \mathrm{x}\mathrm{dy}=\frac{\mathrm{d}}{\mathrm{a}}(\mathrm{dx}) \Rightarrow \frac{1}{\mathrm{dc}}=\frac{\mathrm{y}}{\mathrm{K} \varepsilon_{0} \mathrm{adx}}+\frac{(\mathrm{d}-\mathrm{y})}{\varepsilon_{0} \mathrm{adx}}\frac{1}{d c}=\frac{y}{\varepsilon_{0} a b x}\left(\frac{y}{k}+d-y\right)\int \mathrm{dc}=\int \frac{\varepsilon_{0} \mathrm{adx}}{\frac{\mathrm{y}}{\mathrm{k}}+\mathrm{d}-\mathrm{y}}$or,$c=\varepsilon_{0} a \cdot \frac{a}{d} \int_{0}^{d} \frac{d y}{d+y\left(\frac{1}{k}-1\right)}=\frac{\varepsilon_{0} \mathrm{a}^{2}}{\left(\frac{1}{\mathrm{k}}-1\right) \mathrm{d}}\left[\ell \mathrm{n}\left(\mathrm{d}+\mathrm{y}\left(\frac{1}{\mathrm{k}}-1\right)\right)\right]_{0}^{\mathrm{d}}=\frac{\mathrm{k} \in_{0} \mathrm{a}^{2}}{(1-\mathrm{k}) \mathrm{d}} \ell \mathrm{n}\left(\frac{\mathrm{d}+\mathrm{d}\left(\frac{1}{\mathrm{k}}-1\right)}{\mathrm{d}}\right)=\frac{\mathrm{k} \in_{0} \mathrm{a}^{2}}{(1-\mathrm{k}) \mathrm{d}} \ell \mathrm{n}\left(\frac{1}{\mathrm{k}}\right)=\frac{\mathrm{k} \in_{0} \mathrm{a}^{2} \ell \mathrm{nk}}{(\mathrm{k}-1) \mathrm{d}}\$