(C) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across the plates remains constant $(V = V_0)$.The energy s

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(C) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across the plates remains constant $(V = V_0)$.The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.When a dielectric slab of dielectric constant $K$ is introduced between the plates,the capacitance becomes $C' = KC$.Therefore,the new energy stored is $U' = \frac{1}{2} (KC) V^2 = K U$. Thus,the energy stored becomes $K$ times.Since $Q = CV$,the charge on the plates becomes $Q' = KCV = KQ$. The surface charge density is given by $\sigma = \frac{Q}{A}$. Since $Q$ increases by a factor of $K$,the surface charge density $\sigma' = \frac{KQ}{A} = K\sigma_0$ also increases by a factor of $K$.Therefore,the Reason is incorrect because the surface charge density does not remain constant.

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

Medium

Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.

AIIMS 2008 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

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Electrostatic Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ capacitor has some dielectric between its plates and the capacitor is connected to a $D.C.$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance,the energy stored in it,electric field,charge stored,and the voltage will increase,decrease,or remain constant.

Medium

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The plates of a parallel plate capacitor are charged up to $200 \ V$. $A$ dielectric slab of thickness $4 \ mm$ is inserted between its plates. Then,to maintain the same potential difference between the plates of the capacitor,the distance between the plates is increased by $3.2 \ mm$. The dielectric constant of the dielectric slab is

DifficultAP EAMCET 2004

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$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

MediumAIIMS 2019

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$A$ parallel plate capacitor is made of two plates of length $l$,width $w$ and separated by distance $d$. $A$ dielectric slab (dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{\partial U}{\partial x}$ where $U$ is the energy of the capacitor when the dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$,then the force on the dielectric when it is near the edge is

DifficultJEE MAIN 2014

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$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $k_1 = 3$ and thickness $d/3$,while the other one has a dielectric constant $k_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now . . . . . . $pF$.

MediumAIEEE 2008

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Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.Reason : The surface density of charge on the plate remains constant or unchanged.

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$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

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Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.