A parallel plate capacitor with air as dielectric is charged by a d.c. source to a potential V. Without disconnecting the capacitor from the source, air is replaced by another dielectric medium of dielectric constant K. State with reason, how does
(i) potential difference
(ii) electric field between the plates
(iii) capacity
(iv) charge and
(v) energy stored in the capacitor, change.

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.