Given, a parallel plate capacitor with air as dielectric is charged by d.c source. Later on, without disconnecting the source, air is replaced by another dielectric medium K.
Initial capacitance when, air is the dielectric medium, C_o= $\frac{\mathbf{Q}}{\mathbf{V}}=\frac{{\epsilon }_{\circ }A}{d}$

Capacitance when, dielectric medium of dielectric constant K is introduced, C=$\mathrm{K}\frac{{\mathrm{\epsilon }}_{\circ }\mathrm{A}}{\mathrm{d}}$

(i) Potential difference will remain constant as the capacitor remains connected to the battery. 

(ii) Electric field is given by $\mathbf{E}\mathbf{=}\frac{\mathbf{V}}{\mathbf{d}}$. 
Since, neither the potential difference nor the separation between the plates is changing therefore, the electric field remains unchanged.

(iii) When, dielectric medium is introduced, capacity would increase by a factor of K.
                       So, $\boxed{\mathrm{C} = {\mathrm{KC}}_0.}$ 

(iv) Since, capacitance is increased by a factor of K and potential remains unchanged therefore charge is increased by a factor of K. Additional charge flows from the battery to the plates.

(v)  Energy stored in the capacitor is,
                       $$\begin{gathered} {\mathrm{U}}_0 = \frac{1}{2}{\mathrm{C}}_0{\mathrm{V}}_0^2, \\ \mathrm{U} = \frac{1}{2}\left({\mathrm{KC}}_0\right){\mathrm{V}}^2 \\ ={\mathrm{KU}}_0 \end{gathered}$$.
Therefore, the energy is increased by a factor of K.

Electric potential at a point is defined as the amount of work done, in moving a unit positive charge with zero acceleration from infinity to that point.

Potential energy at a position is equal to the amount of work done to carry the total charge from infinity to that position, against the electrostatic forces.
Thus, potential energy = potential x charge.
Potential energy, is the energy possessed by the charge by virtue of its particular position.