Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
Let's assume contradicting statement that the potential is not same inside the closed equipotential surface. Let the potential just inside the surface is different to that of the surface causing in a potential gradient $\left(\frac{d V}{d r}\right)$. Consequently electric field comes into existence, which is given by as $E=-\frac{d V}{d r}$.
Consequently field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside.
This contradict the original assumption. Hence, the entire volume inside must be equipotential.
A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed . State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.
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 $E=V / d$.
Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity.
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.
Calculate potential energy of a point charge $-q$ placed along the axis due to a charge $+Q$ uniformly distributed along a ring of radius $R$. Sketch PE, as a function of axial distance $z$ from the centre of the ring. Looking at graph, can you see what would happen if $-q$ is displaced slightly from the centre of the ring (along the axis)?
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.
Calculate potential on the axis of a ring due to charge $Q$ uniformly distributed along the ring of radius $R$.
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 $$V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{x^2+a^2}}$$