A parallel plate capacitor having a separation between the plates $\mathrm{d}$, plate area $\mathrm{A}$ and material with dielectric constant $\mathrm{K}$ has capacitance $\mathrm{C}_0$. Now one-third of the material is replaced by another material with dielectric constant $2 \mathrm{~K}$, so that…

Question

A parallel plate capacitor having a separation between the plates $\mathrm{d}$, plate area $\mathrm{A}$ and material with dielectric constant $\mathrm{K}$ has capacitance $\mathrm{C}_0$. Now one-third of the material is replaced by another material with dielectric constant $2 \mathrm{~K}$, so that effectively there are two capacitors one with area $\frac{1}{3} \mathrm{~A}$, dielectric constant $2 \mathrm{~K}$ and another with area $\frac{2}{3} \mathrm{~A}$ and dielectric constant $\mathrm{K}$. If the capacitance of this new capacitor is $\mathrm{C}$ then $\frac{\mathrm{C}}{\mathrm{C}_0}$ is

Aicharya · Accepted Answer

$\mathrm{C}_0=\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C}=\frac{\mathrm{k} \in_0 2}{3 \mathrm{~d}}+\frac{2 \mathrm{k} \in_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}$ $	herefore \frac{\mathrm{C}}{\mathrm{C}_0}=\frac{\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}{\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{4}{3}$

A parallel plate capacitor having a separation between the plates $\mathrm{d}$, plate area $\mathrm{A}$ and…

Solution