(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,
$$\begin{aligned} U & =\frac{1}{4 \pi \varepsilon_0}\left\{\frac{q_d q_d}{r}-\frac{q_u q_d}{r}-\frac{q_u q_d}{r}\right\} \\ & =\frac{9 \times 10^9}{10^{-15}}\left(1.6 \times 10^{-19}\right)^2\left[\left\{(1 / 3)^2-(2 / 3)(1 / 3)-(2 / 3)(1 / 3)\right\}\right] \\ & =2.304 \times 0^{-13}\left\{\frac{1}{9}-\frac{4}{9}\right\}=-7.68 \times 10^{-14} \mathrm{~J} \\ & =4.8 \times 10^5 \mathrm{eV}=0.48 \mathrm{meV}=5.11 \times 10^{-4}\left(m_n c^2\right) \end{aligned}$$
Two metal spheres, one of radius $R$ and the other of radius $2 R$, both have same surface charge density $\sigma$. They are brought in contact and separated. What will be new surface charge densities on them?
The charges on two metal spheres, before coming in contact, are given by
$$\begin{aligned} Q & =\sigma .4 \pi R^2 \\ Q_2 & =\sigma .4 \pi\left(2 R^2\right) \\ & =4\left(\sigma .4 \pi R^2\right)=4 Q_1 \end{aligned}$$
Let the charges on two metal spheres, after coming in contact becomes $Q_1^{\prime}$ and $Q_2^{\prime}$.
Now applying law of conservation of charges is given by
$$\begin{aligned} Q_1^{\prime}+Q_2^{\prime} & =Q_1+Q_2=5 Q_1 \\ & =5\left(\sigma \cdot 4 \pi R^2\right) \end{aligned}$$
After coming in contact, they acquire equal potentials. Therefore, we have
$$\frac{1}{4 \pi \varepsilon_0} \frac{Q_1^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{Q_2^{\prime}}{R}$$
On solving, we get
$$\begin{array}{ll} \therefore & Q_1^{\prime}=\frac{5}{3}\left(\sigma .4 \pi R^2\right) \text { and } Q_2^{\prime}=\frac{10}{3}\left(\sigma .4 \pi R^2\right) \\ \therefore & \sigma_1=5 / 3 \sigma \text { and } \\ \therefore & \sigma_2=\frac{5}{6} \sigma \end{array}$$
In the circuit shown in figure, initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitors? Then $K_1$ was opened and $K_2$ was closed (order is important), what will be the charge on each capacitor now? $[C=1 \propto \mathrm{~F}]$
In the circuit, when initially $K_1$ is closed and $K_2$ is open, the capacitors $C_1$ and $C_2$ acquires potential difference $V_1$ and $V_2$ respectively. So, we have
$$\begin{aligned} &\text { and }\\ &\begin{aligned} & V_1+V_2=E \\ & V_1+V_2=9 V \end{aligned} \end{aligned}$$
Also, in series combination ,
$$\begin{aligned} V & \propto 1 / C \\ V_1: V_2 & =1 / 6: 1 / 3 \end{aligned}$$
On solving
$$\begin{array}{ll} \Rightarrow & V_1=3 V \text { and } V_2=6 \mathrm{~V} \\ \therefore & Q_1=C_1 V_1=6 \mathrm{C} \times 3=18 \propto \mathrm{C} \\ & Q_2=9 \propto \mathrm{C} \text { and } Q_3=0 \end{array}$$
Then, $K_1$ was opened and $K_2$ was closed, the parallel combination of $C_2$ and $C_3$ is in series with $C_1$.
$$Q_2=Q_2^{\prime}+Q_3$$
and considering common potential of parallel combination as $V$, then we have
$$\begin{aligned} C_2 V+C_3 V & =Q_2 \\ \Rightarrow\quad V & =\frac{Q_2}{C_2+C_3}=(3 / 2) \mathrm{V} \\ \text{On solving,}\quad Q_2^{\prime} & =(9 / 2) \propto \mathrm{C} \\ \text{and}\quad Q_3 & =(9 / 2) \propto \mathrm{C} \end{aligned}$$
Calculate potential on the axis of a disc of radius $R$ due to a charge $Q$ uniformly distributed on its surface.
Let the point $P$ lies at a distance $x$ from the centre of the disk and take the plane of the disk to be perpendicular to the $\mathbf{x}$-axis. Let the disc is divided into a number of charged rings as shown in figure.
The electric potential of each ring, of radius $r$ and width $d r$, have charge $d q$ is given by
$$\sigma d A=\sigma 2 \pi r d r$$
and potential is given by
$$d V=\frac{k_e d q}{\sqrt{r^2+x^2}}=\frac{k_e \sigma 2 \pi r d r}{\sqrt{r^2+x^2}}$$
where $k_e=\frac{1}{4 \pi \varepsilon_0}$ the total electric potential at $P$, is given by
$$\begin{aligned} &\begin{aligned} & V=\pi k_e \sigma \int_0^a \frac{2 r d r}{\sqrt{r^2+x^2}}=\pi k_e \sigma \int_0^a\left(r^2+x^2\right)^{-1 / 2} 2 r d r \\ & V=2 \pi k_e \sigma\left[\left(x^2+a^2\right)^{1 / 2}-x\right] \end{aligned}\\ &\begin{aligned} \text{So, we have by substring}\quad k_e & =\frac{1}{4 \pi \varepsilon_0} \\ V & =\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{a^2}\left[\sqrt{x^2+a^2}-x\right] \end{aligned} \end{aligned}$$
Two charges $q_1$ and $q_2$ are placed at $(0,0, d)$ and $(0,0,-d)$ respectively. Find locus of points where the potential is zero.
Let any arbitrary point on the required plane is $(x, y, z)$. The two charges lies on $z$-axis at a separation of $2 d$.
The potential at the point $P$ due to two charges is given by
$$\begin{array}{ll} & \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}+\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}}=0 \\ \therefore & \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}=\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}} \end{array}$$
On squaring and simplifying, we get
$$x^2+y^2+z^2+\left[\frac{\left(q_1 / q_2\right)^2+1}{\left(q_1 / q_2\right)^2-1}\right](2 z d)+d^2-0$$
This is the equation of a sphere with centre at
$$\left(0,0,-2 d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right]\right)$$