(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

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

(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 $C = \frac{Q}{V}$:$C = \frac{Q}{\frac{Q}{A\varepsilon_0} (d - t(1 - \frac{1}{K}))} = \frac{\varepsilon_0 A}{d - t(1 - \frac{1}{K})}$

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

Medium

$A$ 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?

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

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ 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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$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \text{ pF}$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4 \text{ pF}$. The value of $x$ is

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