# A parallel plate capacitor

Question:

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

$\varepsilon(x)=\varepsilon_{0}+k x$, for $\left(0$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for$\left(\frac{d}{2} \leq x \leq d\right)$1.$\left(\varepsilon_{0}+\frac{\mathrm{kd}}{2}\right)^{2 / \mathrm{kA}}$2.$\frac{\mathrm{kA}}{2 \ln \left(\frac{2 \varepsilon_{0}+\mathrm{kd}}{2 \varepsilon_{0}}\right)}$3. 0 4.$\frac{\mathrm{kA}}{2} \ln \left(\frac{2 \varepsilon_{0}}{2 \varepsilon_{0}-\mathrm{kd}}\right)$Correct Option: , 2 Solution: Taking an element of width$\mathrm{dx}$at a distance$\mathrm{x}(\mathrm{x}<\mathrm{d} / 2)$from left plate$\mathrm{dc}=\frac{\left(\varepsilon_{0}+\mathrm{kx}\right) \mathrm{A}}{\mathrm{dx}}$Capacitance of half of the capacitor$\frac{1}{\mathrm{C}}=\int_{0}^{\mathrm{d} / 2} \frac{1}{\mathrm{dc}}=\frac{1}{\mathrm{~A}} \int_{0}^{\mathrm{d} / 2} \frac{\mathrm{dx}}{\varepsilon_{0}+\mathrm{kx}}\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{kA}} \ln \left(\frac{\varepsilon_{0}+\mathrm{kd} / 2}{\varepsilon_{0}}\right)$Capacitance of second half will be same$\mathrm{C}_{\mathrm{cq}}=\frac{\mathrm{C}}{2}=\frac{\mathrm{kA}}{2 \ln \left(\frac{2 \varepsilon_{0}+\mathrm{kd}}{2 \varepsilon_{0}}\right)}\$