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 $$V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{x^2+a^2}}$$
Find the equation of the equipotentials for an infinite cylinder of radius $r_0$ carrying charge of linear density $\lambda$.
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.
Two point charges of magnitude $+q$ and $-q$ are placed at $(-d / 2,0,0)$ and $(d / 2,2,0)$, respectively. Find the equation of the equipotential surface where the potential is zero.
Let the required plane lies at a distance x from the origin as shown in figure.
The potential at the point $P$ due to charges is given by
$$\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left[(x+d / 2)^2+h^2\right]^{1 / 2}}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left[(x-d / 2)^2+h^2\right]^{1 / 2}}$$
If net electric potential is zero, then
$$\begin{aligned} \frac{1}{\left[(x+d / 2)^2+h^2\right]^{1 / 2}} & =\frac{1}{\left[(x-d / 2)+h^2\right]^{1 / 2}} \\ \text{Or}\quad (x-d / 2)^2+h^2 & =(x+d / 2)^2+h^2 \\ \Rightarrow\quad x^2-d x+d^2 / 4 & =x^2+d x+d^2 / 4 \\ \text{Or}\quad 2 d x & =0 \Rightarrow x=0 \end{aligned}$$
The equation of the required plane is $x=0$ i.e., $y-z$ plane.
A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as $\varepsilon=\alpha U$ where $a=2 \mathrm{~V}^{-1}$. A similar capacitor with no dielectric is charged to $U_0=78 \mathrm{~V}$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
Assuming the required final voltage be $U$. If $C$ is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is given by $Q_1=C U$
Since, the capacitor with the dielectric has a capacitance $\varepsilon C$. Hence, the charge on the capacitor is given by
$$Q_2=\varepsilon C U=(\alpha U) C U=\alpha C U$$
The initial charge on the capacitor is given by
$$Q_0=C U_0$$
From the conservation of charges,
$$\begin{aligned} Q_0 & =Q_1+Q_2 \\ \text{Or}\quad C U_0 & =C U+\alpha C U^2 \end{aligned}$$
$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad \alpha U^2+U-U_0=0 \\ & \therefore \quad U=\frac{-1 \pm \sqrt{1+4 \alpha U_0}}{2 \alpha} \end{aligned}\\ &\text { On solving for } U_0=78 \mathrm{~V} \text { and } a=2 / \mathrm{V} \text {, we get }\\ &U=6 \mathrm{~V} \end{aligned}$$
A capacitor is made of two circular plates of radius $R$ each, separated by a distance $d \ll R$. The capacitor is connected to a constant voltage. A thin conducting disc of radius $r \ll R$ and thickness $t \ll r$ is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is $m$.
Assuming initially the disc is in touch with the bottom plate, so the entire plate is a equipotential.
The electric field on the disc, when potential difference $V$ is applied across it, given by
$$E=\frac{V}{d}$$
Let charge $q^{\prime}$ is transferred to the disc during the process,
Therefore by Gauss' theorem,
$$\begin{aligned} \therefore\quad q^{\prime} & =-\varepsilon_0 \frac{V}{d} \pi r^2 \\ \text{Since, Gauss theorem states that}\quad \phi & =\frac{q}{\varepsilon_0} \text { or } q=\frac{\varepsilon_0}{\phi} \\ & =\varepsilon E A=\frac{\varepsilon_0 V}{d} A \end{aligned}$$
The force acting on the disc is
$$-\frac{V}{d} \times q^{\prime}=\varepsilon_0 \frac{V^2}{d^2} \pi r^2$$
If the disc is to be lifted, then
$$\varepsilon_0 \frac{V^2}{d^2} \pi r^2=m g \Rightarrow V=\sqrt{\frac{m g d^2}{\pi \varepsilon_0 r^2}}$$
This is the required expression.