The separation between the plates of a parall | vedclass

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(C) The potential difference $V$ between the plates is the sum of the potential drop across the air gap and the dielectric slab.$V = V_{	ext{air}} +

Vedclass · Accepted Answer

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

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(C) The potential difference $V$ between the plates is the sum of the potential drop across the air gap and the dielectric slab.$V = V_{	ext{air}} + V_{	ext{medium}}$$V = E_0(d - t) + E_{	ext{medium}}t$Since $E_0 = \frac{\sigma}{\varepsilon_0}$ and $E_{	ext{medium}} = \frac{\sigma}{k\varepsilon_0}$,where $\sigma = \frac{Q}{A}$:$V = \frac{\sigma}{\varepsilon_0}(d - t) + \frac{\sigma}{k\varepsilon_0}t$$V = \frac{Q}{A\varepsilon_0} \left( d - t + \frac{t}{k} ight)$$V = \frac{Q}{A\varepsilon_0} \left( d - t(1 - \frac{1}{k}) ight)$The capacitance $C$ is given by $C = \frac{Q}{V}$:$C = \frac{Q}{\frac{Q}{A\varepsilon_0} \left( d - t(1 - \frac{1}{k}) ight)}$$C = \frac{\varepsilon_0 A}{d - t(1 - \frac{1}{k})}$

The separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t$ $(t < d)$ is introduced between the plates,its capacitance becomes

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