The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant $K$ is given by

$$C=\frac{K \varepsilon_0 A}{d} \text {, where signs are as usual. }$$

The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum $K=1$.

After disconnection from battery charge stored will remain the same due to conservation of charge.

The energy stored in an isolated charge capacitor $=\frac{q^2}{2 C}$; as $q$ is constant, energy stored $\propto$ $1 / C$ and $C$ decreases with the removal of dielectric medium, therefore energy stored increases. Since $q$ is constant and $V=q / C$ and $C$ decreases which in turn increases $V$ and therefore $E$ increases as $E=V / d$.

Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.

Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Let us take point $P$ to be at a distance $x$ from the centre of the ring, as shown in figure. The charge element $d q$ is at a distance $x$ from point $P$. Therefore, $V$ can be written as

$$V=k_e \int \frac{d q}{r}=k_e \int \frac{d q}{\sqrt{z^2+a^2}}$$

where, $k=\frac{1}{4 \pi \varepsilon_0}$, since each element $d q$ is at the same distance from point $P$, so we have net potential

$$V=\frac{k_e}{\sqrt{z^2+a^2}} \int d q=\frac{k_e Q}{\sqrt{z^2+a^2}}$$

Considering - $q$ charge at $P$, the potential energy is given by $U=W=q \times$ potential difference

$$U=\frac{k_e Q(-q)}{\sqrt{z^2+a^2}}$$

$$\begin{aligned} \text{or}\quad U & =\frac{1}{4 \pi \varepsilon_0} \frac{-Q q}{\sqrt{z^2+a^2}} \\ & =\frac{1}{4 \pi \varepsilon_0 a} \frac{-Q q}{\sqrt{1+\left(\frac{z}{a}\right)^2}} \end{aligned}$$

This is the required expression.

The variation of potential energy with $z$ is shown in the figure. The charge $-q$ displaced would perform oscillations.

Nothing can be concluded just by looking at the graph.

Let us take point $P$ to be at a distance $x$ from the centre of the ring, as shown in figure. The charge element $d q$ is at a distance $x$ from point $P$. Therefore, $V$ can be written as

$$V=k_e \int \frac{d q}{r}=k_e \int \frac{d q}{\sqrt{x^2+a^2}}$$

where, $k_e=\frac{1}{4 \pi \varepsilon_0}$, since each element $d q$ is at the same distance from point $P$, so we have net potential

$$V=\frac{k_e}{\sqrt{x^2+a^2}} \int d q=\frac{k_e Q}{\sqrt{x^2+a^2}}$$

The net electric potential $$V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{x^2+a^2}}$$

Let the field lines must be radically outward. Draw a cylindrical Gaussian surface of radius $r$ and length $l$. Then, applying Gauss' theorem

$$\begin{aligned} & \int \mathrm{E} \cdot \mathrm{dS}=\frac{1}{\varepsilon_0} \lambda l \\ \text{or}\quad & E_r 2 \pi r l=\frac{1}{\varepsilon_0} \lambda l \Rightarrow E_r=\frac{\lambda}{2 \pi \varepsilon_0 r} \end{aligned}$$

$$\begin{aligned} &\text { Hence, if } r_0 \text { is the radius, }\\ &V(r)-V\left(r_0\right)=-\int_\limits{r_0}^r \mathrm{E} . \mathrm{dl}=\frac{\lambda}{2 \pi \varepsilon_0} \ln \frac{r_0}{r} \end{aligned}$$

Since, $$\int_{r_0}^r \frac{\lambda}{2 \pi \varepsilon_0 r} d r=\frac{\lambda}{2 \pi \varepsilon_0} \int_{r_0}^r \frac{1}{r} d r=\frac{\lambda}{2 \pi \varepsilon_0} \ln \frac{r}{r_0}$$

$$\begin{aligned} \text{For a given V,}\\ \ln \frac{r}{r_0} & =-\frac{2 \pi \varepsilon_0}{\lambda}\left[V(r)-V\left(r_0\right)\right] \\ \Rightarrow\quad r & =r_0 e^{-2 \pi \varepsilon_0 V r_0 / \lambda} e+2 \pi \varepsilon_0 V(r) / \lambda \\ r & =r_0 e^{-2 \pi \varepsilon_0\left[V_{(r)}-V_{\left(r_0\right)}\right) / \lambda} \end{aligned}$$

The equipotential surfaces are cylinders of radius.