(D) $1$. When a dielectric slab of constant $K$ is present,the capacitance is $C_1 = \frac{K \varepsilon_0 A}{d}$.$2$. Since the capacitor remains con

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(D) $1$. When a dielectric slab of constant $K$ is present,the capacitance is $C_1 = \frac{K \varepsilon_0 A}{d}$.$2$. Since the capacitor remains connected to the battery,the potential difference across the plates remains constant,so $V_1 = V_2 = V$.$3$. When the dielectric slab is removed,the new capacitance becomes $C_2 = \frac{\varepsilon_0 A}{d}$.$4$. Since $K > 1$,it is clear that $C_1 > C_2$.$5$. Because the battery is still connected,the potential difference across the plates does not change,thus $V_1 = V_2$.

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

Medium

$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

KCET 2023 All KCET

A
$V_1 = V_2, C_1 < C_2$
B
$V_1 > V_2, C_1 > C_2$
C
$V_1 < V_2, C_1 > C_2$
D
$V_1 = V_2, C_1 > C_2$

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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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The space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance $x$ as per the relation: $K(x) = K_0 + \lambda x$ (where $\lambda$ is a constant). The capacitance $C$ of the capacitor would be related to its vacuum capacitance $C_0$ by the relation:

DifficultJEE MAIN 2014

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If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

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$A$ capacitor stores $60 \ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \ \mu C$ flows through the battery. If the initial capacitance of the capacitor was $2 \ \mu F$,the amount of heat produced when the dielectric is inserted is ... $\mu J$.

Difficult

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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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$A$ parallel plate capacitor has a plate area of $100\, m^{2}$ and a plate separation of $10\, m$. The space between the plates is filled up to a thickness of $5\, m$ with a material of dielectric constant $10$. The resultant capacitance of the system is $'x'\, pF$. Given $\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$,the value of $'x'$ to the nearest integer is:

DifficultJEE MAIN 2021

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If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

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