d
(d) Area of the given metallic plate \(A = \pi r^2\)
Area of the dielectric plate \(A' = \pi \,{\left( {\frac{r}{2}} \right)^2} = \frac{A}{4}\)
Uncovered area of the metallic plates
\( = A - \frac{A}{4} = \frac{{3A}}{4}\)
The given situation is equivalent to a parallel combination of two capacitor. One capacitor \((C')\) is filled with a dielectric medium \((K = 6)\) having area \(\frac{A}{4}\) while the other capacitor \((C'')\) is air filled having area \(\frac{{3A}}{4}\)
Hence
\( = \frac{{{\varepsilon _0}A}}{d}\left( {\frac{K}{4} + \frac{3}{4}} \right)\)\( = \frac{{{\varepsilon _0}A}}{d}\,\left( {\frac{6}{4} + \frac{3}{4}} \right) = \frac{9}{4}\,C\)