(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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Electric Potential and Capacitance

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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?

Similar Questions

Consider two uncharged capacitors of equal capacitance $200 \text{ pF}$. One of them is charged by a $100 \text{ V}$ supply and disconnected. Now, this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

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:

$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another identical uncharged capacitor is connected in parallel with it. The total energy of the system of capacitors is . . . . . . .

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