Question

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates.Find out the expression for its capacitance when the slab is inserted between plates of capacitor.

Answer 1

$$\begin{gathered} \mathrm{Capacitance}\mathrm{of}\mathrm{parallel}\mathrm{plate}\mathrm{air}\mathrm{capacitor}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{C}}_0=\frac{{\in }_0\mathrm{A}}{\mathrm{d}} \\ \mathrm{Capacitance}\mathrm{of}\mathrm{capacitor}\mathrm{when}\mathrm{its}\mathrm{plates}\mathrm{are}\mathrm{partly}\mathrm{filled}\mathrm{with}\mathrm{dielectric}\mathrm{of}\mathrm{thickness}\mathrm{t}\mathrm{and}\mathrm{same}\mathrm{area}\mathrm{as}\mathrm{the}\mathrm{plates}, \\ \mathrm{C}=\frac{{\in }_0\mathrm{A}}{\mathrm{d}-\mathrm{t}(1-{\displaystyle \frac{1}{\mathrm{K}}})} \\ \mathrm{Here}, \\ \mathrm{t}=\frac{\mathrm{d}}{2} \\ \mathrm{So},\mathrm{C}=\frac{{\in }_0\mathrm{A}}{\mathrm{d}-{\displaystyle \frac{\mathrm{d}}{2}}(1-{\displaystyle \frac{1}{\mathrm{K}}})} \\ =\frac{{\in }_0\mathrm{A}}{\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2\mathrm{K}}}=\frac{{\in }_0\mathrm{A}}{{\displaystyle \frac{\mathrm{dK}+\mathrm{d}}{2\mathrm{K}}}} \\ =\frac{{\in }_0\mathrm{A}\times 2\mathrm{K}}{\mathrm{d}(\mathrm{K}+1)}=\frac{2\mathrm{K}}{\mathrm{K}+1}\frac{{\in }_0\mathrm{A}}{\mathrm{d}} \\ \mathrm{Or}\mathrm{C}=\frac{2\mathrm{K}}{\mathrm{K}+1}{\mathrm{C}}_0 \\ \mathrm{Thus},\mathrm{capacitance}\mathrm{becomes}\frac{2\mathrm{K}}{\mathrm{K}+1}\mathrm{times}\mathrm{of}\mathrm{initial}\mathrm{capacitance}. \end{gathered}$$