Two parallel plate capacitors of $8 \mu F$ each are connected in parallel to a $10 \ V$ battery. The plate separation in one of the capacitors is reduced to $40 \%$ of its initial value. The increase in the total charge stored on the capacitors is

In the steady state,what is the potential difference between points $A$ and $B$ in volts? (All capacitors are in $\mu F$.)

$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \; \mu C$ and $-2 \; \mu C$ (with no external field) placed at $(-9 \; cm, 0, 0)$ and $(9 \; cm, 0, 0)$ respectively.
$(b)$ How much work is required to separate the two charges infinitely away from each other?
$(c)$ Suppose that the same system of charges is now placed in an external electric field $E = A(1/r^2)$; $A = 9 \times 10^5 \; C \cdot m^{-2}$. What would the electrostatic energy of the configuration be?

$A$ parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $E_2 / E_1$ is

The figure shows a diagonal symmetric arrangement of capacitors and a battery. If the potential of $C$ is zero,then

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is: