(D) When a dielectric slab is inserted into a parallel plate capacitor while the battery remains connected,the potential difference $V$ across the pla

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(D) When a dielectric slab is inserted into a parallel plate capacitor while the battery remains connected,the potential difference $V$ across the plates remains constant,i.e.,$V = {V_0}$.Since the capacitance $C$ increases by a factor of $K$ (dielectric constant),where $C = KC_0$,the charge $Q$ on the plates increases according to $Q = CV = K C_0 V_0 = K Q_0$. Thus,$Q > {Q_0}$.The electric field $E$ between the plates is given by $E = V/d$. Since $V$ and $d$ remain constant,$E = {E_0}$.The energy stored in the capacitor is $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ remains constant,$U$ increases,i.e.,$U > {U_0}$.Therefore,both statements $(a)$ and $(b)$ are correct.

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

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Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as:

IIT 1985 All IIT

A
$Q > {Q_0}$
B
$U > {U_0}$
C
$E > {E_0}$
D
Both $(a)$ and $(b)$

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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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IIT 1985 Paper All IIT years →

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.

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What will be the capacity of a parallel-plate capacitor when the half of the parallel space between the plates is filled by a material of dielectric constant $\varepsilon_r$? Assume that the capacity of the capacitor in air is $C$.

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$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

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$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The distance between the plates is $d$. Now,the space between the plates is filled with two dielectrics. One dielectric has a dielectric constant $K_1 = 3$ and thickness $d/3$,while the other has a dielectric constant $K_2 = 6$ and thickness $2d/3$. Find the new capacitance of the capacitor in $pF$.

Medium

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$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

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What will be the capacity of a parallel-plate capacitor when the half of the parallel space between the plates is filled by a material of dielectric constant $\varepsilon_r$? Assume that the capacity of the capacitor in air is $C$.

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

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