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.
(a) In a quark model of elementary particles, a neutron is made of one up quarks [charge $(2 / 3) e$ ] and two down quarks [charges - (1/3)e]. Assume that they have a triangle configuration with side length of the order of $10^{-15} \mathrm{~m}$. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV .
(b) Repeat above exercise for a proton which is made of two up and one down quark.
This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of PE of each pair. So,
$$\begin{aligned} U & =\frac{1}{4 \pi \varepsilon_0}\left\{\frac{q_d q_d}{r}-\frac{q_u q_d}{r}-\frac{q_u q_d}{r}\right\} \\ & =\frac{9 \times 10^9}{10^{-15}}\left(1.6 \times 10^{-19}\right)^2\left[\left\{(1 / 3)^2-(2 / 3)(1 / 3)-(2 / 3)(1 / 3)\right\}\right] \\ & =2.304 \times 0^{-13}\left\{\frac{1}{9}-\frac{4}{9}\right\}=-7.68 \times 10^{-14} \mathrm{~J} \\ & =4.8 \times 10^5 \mathrm{eV}=0.48 \mathrm{meV}=5.11 \times 10^{-4}\left(m_n c^2\right) \end{aligned}$$