(B) The capacitors $C_1$ and $C_2$ are connected in parallel to the battery. Therefore,the potential difference across both capacitors is the same,i.e

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(B) The capacitors $C_1$ and $C_2$ are connected in parallel to the battery. Therefore,the potential difference across both capacitors is the same,i.e.,$V_1 = V_2 = V$.The charge on a capacitor is given by the formula $Q = CV$.For capacitor $C_1$,the charge is $Q_1 = C_1 V$.For capacitor $C_2$,the charge is $Q_2 = C_2 V$.Since the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K > 1$,its capacitance increases such that $C_2 = K C_1$,which implies $C_2 > C_1$.Since $V$ is constant and $C_2 > C_1$,it follows that $C_2 V > C_1 V$,which means $Q_2 > Q_1$ or $Q_1 < Q_2$.

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

Easy

Two capacitors $C_1$ and $C_2$ are connected to a battery as shown in the figure. The space between the plates of $C_1$ is filled with air,and the space between the plates of $C_2$ is filled with a dielectric material. Which of the following is true regarding the charges $Q_1$ and $Q_2$ on the capacitors?

A
$Q_1 > Q_2$
B
$Q_1 < Q_2$
C
$Q_1 = Q_2$
D
None of the above

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

DifficultJEE MAIN 2022

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The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

EasyMHT CET 2020

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$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced,is

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$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

Medium

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$A$ $60 \ \mu F$ parallel plate capacitor whose plates are separated by $6 \ mm$ is charged to $250 \ V$,and then the charging source is removed. When a slab of dielectric constant $5$ and thickness $3 \ mm$ is placed between the plates,find the change in the potential difference across the capacitor (in $V$)?

DifficultAP EAMCET 2021

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$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

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