(D) Let the capacitance of the air-filled capacitor be $C$. Its charge is $Q = CV$.Since the capacitors are identical,the capacitance of the dielectri

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(D) Let the capacitance of the air-filled capacitor be $C$. Its charge is $Q = CV$.Since the capacitors are identical,the capacitance of the dielectric-filled capacitor is $C' = KC$.When the two capacitors are connected in parallel,the total charge $Q_{total}$ is conserved.$Q_{total} = Q_1 + Q_2 = 0 + CV = CV$.The equivalent capacitance of the parallel combination is $C_{eq} = C + KC = C(1+K)$.The common potential $V'$ is given by the formula $V' = \frac{Q_{total}}{C_{eq}}$.Substituting the values,we get $V' = \frac{CV}{C(1+K)}$.Therefore,$V' = \frac{V}{1+K}$.

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

Medium

There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

A
$\frac{V}{K-1}$
B
$\frac{K V}{K+1}$
C
$\frac{K V}{K-1}$
D
$\frac{V}{K+1}$

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Class 12 Questions

Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

$A$ capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

MediumWBJEE 2013

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In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?

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$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

EasyAP EAMCET 2001

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$A$ capacitor of capacitance $0.2 \, \mu F$ is charged to a potential of $600 \, V$. After removing the battery,it is connected in parallel to an uncharged capacitor of $1.0 \, \mu F$. The final potential across the capacitors will be.........$V$.

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$A$ $2\,\mu F$ capacitor is charged to a potential of $10\,V$. Another $4\,\mu F$ capacitor is charged to a potential of $20\,V$. The two capacitors are then connected in a single loop,with the positive plate of one connected to the negative plate of the other. What heat is evolved in the circuit in $\mu J$?

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There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

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$A$ capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

$A$ capacitor of capacitance $0.2 \, \mu F$ is charged to a potential of $600 \, V$. After removing the battery,it is connected in parallel to an uncharged capacitor of $1.0 \, \mu F$. The final potential across the capacitors will be.........$V$.

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