(C) The capacitance $C$ of a parallel plate capacitor with multiple dielectric slabs is given by the formula:$C = \frac{{\varepsilon _0 A}}{{\sum \fra

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$\frac{5000}{7}{\varepsilon _0}A$

(C) The capacitance $C$ of a parallel plate capacitor with multiple dielectric slabs is given by the formula:$C = \frac{{\varepsilon _0 A}}{{\sum \frac{t_i}{k_i}}}$Given:$t_1 = 6\,mm = 6 \times 10^{-3}\,m$,$k_1 = 10$$t_2 = 4\,mm = 4 \times 10^{-3}\,m$,$k_2 = 5$Substituting the values:$C = \frac{{\varepsilon _0 A}}{{\frac{6 \times 10^{-3}}{10} + \frac{4 \times 10^{-3}}{5}}}$$C = \frac{{\varepsilon _0 A}}{{0.6 \times 10^{-3} + 0.8 \times 10^{-3}}}$$C = \frac{{\varepsilon _0 A}}{{1.4 \times 10^{-3}}}$$C = \frac{1000}{1.4} \varepsilon _0 A = \frac{10000}{14} \varepsilon _0 A = \frac{5000}{7} \varepsilon _0 A$

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

Medium

The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is

A
$\frac{12}{35}{\varepsilon _0}A$
B
$\frac{2}{3}{\varepsilon _0}A$
C
$\frac{5000}{7}{\varepsilon _0}A$
D
$1500\,{\varepsilon _0}A$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ capacitor has capacity $C$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material of dielectric constant $K$ having an area equal to that of the plates but thickness $\frac{d}{2}$ is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be:

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$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in

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In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

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$A$ parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now,an insulating slab of same area but thickness $t = d/2$ is inserted between the plates as shown in the figure,having dielectric constant $K = 4$. The ratio of the new capacitance to its original capacitance will be:

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The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

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The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is

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$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in

In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

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