(C) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\varepsilon_0 A}{d} = 8 \;pF$. When the dist

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(C) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\varepsilon_0 A}{d} = 8 \;pF$. When the distance between the plates is reduced to $d' = \frac{d}{2}$ and the space is filled with a dielectric of constant $k = 6$,the new capacitance $C'$ is given by: $C' = \frac{k \varepsilon_0 A}{d'} = \frac{6 \varepsilon_0 A}{d/2} = 12 \left( \frac{\varepsilon_0 A}{d} \right)$. Substituting the initial value $C = 8 \;pF$: $C' = 12 \times 8 \;pF = 96 \;pF$. Thus,the new capacitance is $96 \;pF$.

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

Easy

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

A
$64$
B
$32$
C
$96$
D
$128$

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Class 12 Questions

Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor with air between the plates has a capacitance $C$. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant $6$,then the capacitance will become

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When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,which of the following occurs during this process?

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$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

Medium

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The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

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Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

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$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

Similar Questions

$A$ parallel plate capacitor with air between the plates has a capacitance $C$. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant $6$,then the capacitance will become

When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,which of the following occurs during this process?

$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

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