(D) For a capacitor disconnected from the battery,the charge $Q$ remains constant.Initial state: $K_1 = 3$,$Q_1 = Q_0$,$V_1 = V_0 = Q_0/C_1$,$E_1 = E_

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(D) For a capacitor disconnected from the battery,the charge $Q$ remains constant.Initial state: $K_1 = 3$,$Q_1 = Q_0$,$V_1 = V_0 = Q_0/C_1$,$E_1 = E_0 = V_0/d$.New state: $K_2 = 9$,$Q_2 = Q_0$ (since the capacitor is isolated).The new capacitance is $C_2 = K_2 C_0 = 9 C_0$,where $C_0$ is the vacuum capacitance.The new voltage is $V_2 = Q_0 / C_2 = Q_0 / (9 C_0) = (Q_0 / 3 C_0) / 3 = V_0 / 3$.The new electric field is $E_2 = V_2 / d = (V_0 / 3) / d = E_0 / 3$.Thus,the new values are $Q_0, V_0/3, E_0/3$.

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

Medium

When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

A
$3Q_0, 3V_0, 3E_0$
B
$Q_0, 3V_0, 3E_0$
C
$Q_0, V_0/3, 3E_0$
D
$Q_0, V_0/3, E_0/3$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

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Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when a dielectric fills the whole space between the plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when a dielectric fills the whole space between the plates.

EasyAIIMS 2009

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$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

MediumTS EAMCET 2020

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If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

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$A$ capacitor of capacitance $20 \ \mu F$ has a distance of $2 \ mm$ between its plates. If a dielectric slab of thickness $1 \ mm$ and dielectric constant $K = 2$ is inserted between the plates,the new capacitance will be ..... $\mu F$.

Medium

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When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

Similar Questions

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

$A$ capacitor of capacitance $20 \ \mu F$ has a distance of $2 \ mm$ between its plates. If a dielectric slab of thickness $1 \ mm$ and dielectric constant $K = 2$ is inserted between the plates,the new capacitance will be ..... $\mu F$.

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