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(D) The initial capacitance of the parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$.When a dielectric slab of radius $r

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$9C/4$

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(D) The initial capacitance of the parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$.When a dielectric slab of radius $r/2$ and thickness $d$ is inserted,the area covered by the dielectric is $A' = \pi (r/2)^2 = A/4$.The remaining area of the capacitor is filled with air,which is $A'' = A - A/4 = 3A/4$.This arrangement is equivalent to two capacitors connected in parallel.The capacitance of the part filled with the dielectric is $C' = \frac{K \varepsilon_0 A'}{d} = \frac{6 \varepsilon_0 (A/4)}{d} = \frac{6}{4} C = \frac{3}{2} C$.The capacitance of the air-filled part is $C'' = \frac{\varepsilon_0 A''}{d} = \frac{\varepsilon_0 (3A/4)}{d} = \frac{3}{4} C$.The total capacitance is $C_{eq} = C' + C'' = \frac{3}{2} C + \frac{3}{4} C = \frac{6+3}{4} C = \frac{9}{4} C$.

Two metallic plates of radius $r$ are placed at a distance $d$ apart,and the capacitance is $C$. If a plate of radius $r/2$ and thickness $d$ with a dielectric constant $6$ is placed between the plates of the capacitor,then its new capacitance will be

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