(D) Total charge $Q = Q_1 + Q_2 = 80 \ \mu C + 40 \ \mu C = 120 \ \mu C$.When connected by a conducting wire,the spheres reach a common potential $V$.

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(D) Total charge $Q = Q_1 + Q_2 = 80 \ \mu C + 40 \ \mu C = 120 \ \mu C$.When connected by a conducting wire,the spheres reach a common potential $V$.The new charges $Q_1'$ and $Q_2'$ on the spheres of radii $r_1 = 4 \ cm$ and $r_2 = 6 \ cm$ are given by:$Q_1' = Q \left( \frac{r_1}{r_1 + r_2} \right) = 120 \left( \frac{4}{4 + 6} \right) = 120 \times 0.4 = 48 \ \mu C$.The initial charge on the $4 \ cm$ sphere was $80 \ \mu C$.The charge that flows from the $4 \ cm$ sphere to the $6 \ cm$ sphere is $\Delta Q = Q_1 - Q_1' = 80 \ \mu C - 48 \ \mu C = 32 \ \mu C$.

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

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$A$ sphere of radius $4 \ cm$ has a charge of $80 \ \mu C$ and a sphere of radius $6 \ cm$ has a charge of $40 \ \mu C$. If they are connected by a conducting wire,how much charge will flow from the $4 \ cm$ sphere to the $6 \ cm$ sphere?

A
$20 \ \mu C$
B
$16 \ \mu C$
C
$1.694 \times 10^9 \ m^2$
D
$32 \ \mu C$

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Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

The energy stored in a capacitor is $U$. If an uncharged capacitor of the same capacity is connected in parallel with it,then the energy stored in each capacitor is:

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Two capacitors $C_1$ and $C_2$ are connected in parallel. If a charge $Q$ is given to the combination,the charge gets shared. Then the ratio of charge on $C_1$ to charge on $C_2$ is . . . . . . .

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$A$ capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

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Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

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

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$A$ sphere of radius $4 \ cm$ has a charge of $80 \ \mu C$ and a sphere of radius $6 \ cm$ has a charge of $40 \ \mu C$. If they are connected by a conducting wire,how much charge will flow from the $4 \ cm$ sphere to the $6 \ cm$ sphere?

Similar Questions

The energy stored in a capacitor is $U$. If an uncharged capacitor of the same capacity is connected in parallel with it,then the energy stored in each capacitor is:

Two capacitors $C_1$ and $C_2$ are connected in parallel. If a charge $Q$ is given to the combination,the charge gets shared. Then the ratio of charge on $C_1$ to charge on $C_2$ is . . . . . . .

$A$ capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

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