A parallel plate capacitor has plates of area | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{{\varepsilon _0}A}{{d - t(1 - 1/K)}}$

Vedclass · Answer

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

$A$ parallel plate capacitor has plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?

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