(B) $1$. When the capacitor is isolated,the charge $Q$ on the plates remains constant.$2$. The capacitance $C$ of a parallel plate capacitor is given

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Potential difference across the plates decreases.

(B) $1$. When the capacitor is isolated,the charge $Q$ on the plates remains constant.$2$. The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$. Since the separation $d$ is increased,the capacitance $C$ decreases.$3$. The potential difference $V$ across the plates is given by $V = \frac{Q}{C}$.$4$. Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.$5$. Therefore,the statement that the potential difference decreases is incorrect.

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

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$A$ parallel plate capacitor is charged and then isolated. If the separation between the plates is increased,which one of the following statements is $NOT$ correct?

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A
Charge remains constant after it is isolated.
B
Potential difference across the plates decreases.
C
Potential difference across the plates increases.
D
Capacitance of capacitor decreases.

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

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$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

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$A$ parallel plate capacitor is charged and then isolated. If the separation between the plates is increased,which one of the following statements is $NOT$ correct?

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$A$ parallel plate air capacitor has capacity $C$ farad,potential $V$ volt,and energy $E$ joule. When the gap between the plates is completely filled with a dielectric material of dielectric constant $K > 1$,what happens to the potential $V$ and energy $E$?

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: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

$A$ parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

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