$A$ capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2C$. The final charges on the capacitors are

$A$ capacitor of capacitance $C_1 = 10 \mu F$ is charged using a $9 \text{ V}$ battery. It is then removed from the battery and connected to another uncharged capacitor $C_2 = 20 \mu F$ as shown in the figure. The charge on $C_2$ after equilibrium is reached is:

$A$ $4 \mu F$ capacitor is charged by a $200 \ V$ battery. It is then disconnected from the supply and connected to another uncharged $2 \mu F$ capacitor. During the process,the loss of energy (in $J$) is:

$A$ $2 \mu F$ capacitor is charged to a potential $V_1 = 10 \ V$. Another $4 \mu F$ capacitor is charged to a potential $V_2 = 20 \ V$. 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 $\mu J$)

$A$ 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 charged spheres of radii $a$ and $b$ are connected by a wire. The ratio of the electric fields on their surfaces $E_a/E_b$ is: