A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta=\frac{ d }{ N }$. The dielectric constant of the $m ^{\text {th }}$ layer is $K _{ m }= K \left(1+\frac{ m }{ N }\right)$. For a very large $N \left(>10^3\right)$, the capacitance $C$ is $\alpha\left(\frac{ K \varepsilon_0 A }{ d \;ln 2}\right)$. The value of $\alpha$ will be. . . . . . . .
[ $\epsilon_0$ is the permittivity of free space]
A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta=\frac{ d }{ N }$. The dielectric constant of the $m ^{\text {th }}$ layer is $K _{ m }= K \left(1+\frac{ m }{ N }\right)$. For a very large $N \left(>10^3\right)$, the capacitance $C$ is $\alpha\left(\frac{ K \varepsilon_0 A }{ d \;ln 2}\right)$. The value of $\alpha$ will be. . . . . . . .
[ $\epsilon_0$ is the permittivity of free space]
- [IIT 2019]
- A
$1$
- B
$3$
- C
$5$
- D
$6$
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[permittivity of free space $ = 9 \times 10^{-12}\ Fm^{-1}$]
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Assertion : The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates.
Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.
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Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.
- [AIIMS 2009]
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