In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

$A$ parallel plate capacitor is made of two plates of length $l$,width $w$ and separated by distance $d$. $A$ dielectric slab (dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{\partial U}{\partial x}$ where $U$ is the energy of the capacitor when the dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$,then the force on the dielectric when it is near the edge is

$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

$A$ capacitor of capacitance $20 \ \mu F$ has a distance of $2 \ mm$ between its plates. If a dielectric slab of thickness $1 \ mm$ and dielectric constant $K = 2$ is inserted between the plates,the new capacitance will be ..... $\mu F$.

$A$ parallel plate capacitor is made of two square parallel plates of area $A$,and separated by a distance $d << \sqrt{A}$. The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. $A$ square metal conducting slab also with area $A$ but thickness $d/2$ is then fully inserted between the plates,so that it is always parallel to the plates. How much work has been done on the metal slab by an external agent while it is being inserted?