(C) Initial charge on the $6\ \mu F$ capacitor is $Q = C_1 V_0 = 6\ \mu F \times 100\, V = 600\ \mu C$.When connected in parallel,both capacitors reac

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(C) Initial charge on the $6\ \mu F$ capacitor is $Q = C_1 V_0 = 6\ \mu F \times 100\, V = 600\ \mu C$.When connected in parallel,both capacitors reach a common potential $V$.The charge on the capacitors will be $q_1 = C_1 V$ and $q_2 = C_2 V$.The ratio of charges is $\frac{q_1}{q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2} = \frac{6}{14}$.By conservation of charge,$q_1 + q_2 = Q = 600\ \mu C$.$C_1 V + C_2 V = 600\ \mu C \Rightarrow V(6 + 14) = 600$.$V = \frac{600}{20} = 30\, V$.Thus,the ratio is $6/14$ and the potential is $30\, V$.

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

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$A$ capacitor of capacitance $6\ \mu F$ is charged to $100\, V$. It is then connected to another uncharged capacitor of capacitance $14\ \mu F$. Find the ratio of charges on the $6\ \mu F$ and $14\ \mu F$ capacitors and the potential across the $6\ \mu F$ capacitor.

A
$6/14$ and $50\, V$
B
$14/6$ and $30\, V$
C
$6/14$ and $30\, V$
D
$14/6$ and $0\, V$

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

In the following figure,$C_1=5 \mu F$,$C_2=C_3=10 \mu F$ and $\varepsilon=20 \text{ V}$. Initially,the switch $S$ is connected to point $A$ until capacitor $C_1$ is fully charged. Afterwards,the switch is thrown to the left side and connected to point $B$. The charge on capacitor $C_3$ after equilibrium is reached will be (in $\mu C$)

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$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a capacitor of capacity $C_2$ as shown in the figure. The ratio of the total energy before and after the connection of switch $S$ is:

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In the circuit shown in the figure,initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitor? Then $K_1$ was opened and $K_2$ was closed (order is important),what will be the charge on each capacitor now? [Given: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]

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$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

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Two identical capacitors are joined in parallel,charged to a potential $V$ and then separated and then connected in series,i.e.,the positive plate of one is connected to the negative plate of the other. What happens to the system?

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$A$ capacitor of capacitance $6\ \mu F$ is charged to $100\, V$. It is then connected to another uncharged capacitor of capacitance $14\ \mu F$. Find the ratio of charges on the $6\ \mu F$ and $14\ \mu F$ capacitors and the potential across the $6\ \mu F$ capacitor.

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$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a capacitor of capacity $C_2$ as shown in the figure. The ratio of the total energy before and after the connection of switch $S$ is:

$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

Two identical capacitors are joined in parallel,charged to a potential $V$ and then separated and then connected in series,i.e.,the positive plate of one is connected to the negative plate of the other. What happens to the system?

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