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(A) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula:$C =

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$\frac{\varepsilon_0 A}{d - t(1 - 1/K)}$

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(A) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula:$C = \frac{\varepsilon_0 A}{d - t + t/K}$This can be rearranged as:$C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$Here,the effective distance between the plates is reduced due to the presence of the dielectric slab. The potential difference $V$ across the plates is the sum of the potential drop across the air gap $(d-t)$ and the dielectric slab $t$:$V = E_{	ext{air}}(d-t) + E_{	ext{medium}}(t)$Since $E_{	ext{air}} = \frac{\sigma}{\varepsilon_0}$ and $E_{	ext{medium}} = \frac{\sigma}{K\varepsilon_0}$,where $\sigma = \frac{Q}{A}$:$V = \frac{Q}{A\varepsilon_0}(d-t) + \frac{Q}{A\varepsilon_0 K}(t) = \frac{Q}{A\varepsilon_0} [d - t + t/K] = \frac{Q}{A\varepsilon_0} [d - t(1 - 1/K)]$Using $C = \frac{Q}{V}$,we get $C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$.

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

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