Putting a dielectric substance between two plates of condenser, capacity, potential and potential energy respectively

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu \,F$

The distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $ x=0$ and negatively charged plate is at $x=5d$. Two slabs one of conductor and other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like :

A capacitor when filled with a dielectric $K = 3$ has charge ${Q_0}$, voltage ${V_0}$ and field ${E_0}$. If the dielectric is replaced with another one having $K = 9$ the new values of charge, voltage and field will be respectively

A parallel plate capacitor is made of two square plates of side $a$, separated by a distance $d\,(d  < < a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$, as shown in the figure. Capacitance of this capacitor is

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in