$A$ parallel plate capacitor has charge $5 \times 10^{-6} \ C$. $A$ dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \ C$, then the dielectric constant of the slab is . . . . . . .

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ and of the same thickness are inserted in a parallel plate capacitor. Given $K_1 = 2K_2$. If the potential differences across the slabs are $V_1$ and $V_2$ respectively,then:

$A$ capacitor is half-filled with a dielectric $(K=2)$ as shown in Figure $A$. If the same capacitor is to be filled with the same dielectric as shown in Figure $B$, what would be the thickness $t$ of the dielectric so that the capacitor still has the same capacity?

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

$A$ parallel plate capacitor with air between the plates has a capacitance of $1.0 \text{ pF}$. If the distance between the plates is doubled and the space between them is filled with a dielectric substance, the capacitance becomes $2.0 \text{ pF}$. Then the value of the dielectric constant of the dielectric substance is . . . . . . .

$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)