(D) Let the initial capacitance be $C_0$ and the initial potential difference be $V_0$. When a dielectric slab of thickness $d$ (equal to the plate se

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(D) Let the initial capacitance be $C_0$ and the initial potential difference be $V_0$. When a dielectric slab of thickness $d$ (equal to the plate separation) is introduced,the new capacitance $C$ becomes $K C_0$,where $K$ is the dielectric constant.Since the charge $Q$ on the capacitor remains constant $(Q = C_0 V_0 = C V)$,we have $V = Q/C = (C_0 V_0) / (K C_0) = V_0 / K$.Given that the potential difference reduces to $1/8$ of the original value,we have $V = V_0 / 8$.Comparing the two expressions,$V_0 / K = V_0 / 8$,which gives $K = 8$.

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

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The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is

A
$1.6$
B
$40$
C
$5$
D
$8$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively:

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The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $K=4$. Then the new force of repulsion between those two charges becomes:

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When a dielectric of constant $K = 2$ is filled in a capacitor,its capacitance becomes $C$. If the oil is removed,its capacitance will become ......

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$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

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The voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of $10^6\,V/m$. The plate area is $10^{-4}\,m^2$. What is the dielectric constant if the capacitance is $15\,pF$? (Given $\epsilon_0 = 8.86 \times 10^{-12}\,C^2/Nm^2$)

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The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is

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$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively:

The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $K=4$. Then the new force of repulsion between those two charges becomes:

When a dielectric of constant $K = 2$ is filled in a capacitor,its capacitance becomes $C$. If the oil is removed,its capacitance will become ......

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

The voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of $10^6\,V/m$. The plate area is $10^{-4}\,m^2$. What is the dielectric constant if the capacitance is $15\,pF$? (Given $\epsilon_0 = 8.86 \times 10^{-12}\,C^2/Nm^2$)

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