In 1959 Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge. Suppose that the universe is made up of hydrogen atoms with a number density $N$, which is maintained a constant. Let the charge on the proton be $e_p=-(1+y) e$ where $e$ is the electronic charge.
(a) Find the critical value of $y$ such that expansion may start.
(b) Show that the velocity of expansion is proportional to the distance from the centre.
(a) Let us suppose that universe is a perfect sphere of radius $R$ and its constituent hydrogen atoms are distributed uniformly in the sphere.
As hydrogen atom contains one proton and one electron, charge on each hydrogen atom.
$$e_H=e_P+e=-(1+Y) e+e=-Y e=(Y e)$$
If $E$ is electric field intensity at distance $R$, on the surface of the sphere, then according to Gauss' theorem,
$$\begin{aligned} \oint \text { E.ds } & =\frac{q}{\varepsilon_0} \text { i.e., } E\left(4 \pi R^2\right)=\frac{4}{3} \frac{\pi R^3 N \mid Y_{e \mid}}{\varepsilon_0} \\ E & =\frac{1}{3} \frac{N|Y e| R}{\varepsilon_0}\quad\text{... (i)} \end{aligned}$$
Now, suppose, mass of each hydrogen atom $\simeq m_P=$ Mass of a proton, $G_R=$ gravitational field at distance $R$ on the sphere.
$$\begin{aligned} \text{Then,}\quad-4 \pi R^2 G_R & =4 \pi G m_P\left(\frac{4}{3} \pi R^3\right) N \\ \Rightarrow\quad G_R & =\frac{-4}{3} \pi G m_P N R\quad\text{.... (ii)} \end{aligned}$$
$\therefore$ Gravitational force on this atom is $F_G=m_P \times G_R=\frac{-4 \pi}{3} G m_P^2 N R\quad\text{.... (iii)}$
Coulomb force on hydrogen atom at $R$ is $F_C=(Y e) E=\frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0}\quad$ [from Eq. (i)]
Now, to start expansion $F_C>F_G$ and critical value of $Y$ to start expansion would be when
$$\begin{aligned} F_C & =F_G \\ \Rightarrow \quad \frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0} & =\frac{4 \pi}{3} G m_P^2 N R \\ \Rightarrow \quad Y^2 & =\left(4 \pi \varepsilon_0\right) G\left(\frac{m_P}{e}\right)^2 \\ & =\frac{1}{9 \times 10^9} \times\left(6.67 \times 10^{-11}\right)\left(\frac{\left(1.66 \times 10^{-27}\right)^2}{\left(1.6 \times 10^{-19}\right)^2}\right)=79.8 \times 10^{-38} \\ \Rightarrow \quad Y & =\sqrt{79.8 \times 10^{-38}}=8.9 \times 10^{-19} \simeq 10^{-18} \end{aligned}$$
Thus, $10^{-18}$ is the required critical value of $Y$ corresponding to which expansion of universe would start.
(b) Net force experience by the hydrogen atom is given by
$$F=F_C-F_G=\frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N R$$
If acceleration of hydrogen atom is represent by $d^2 R / d t^2$, then
$$\begin{aligned} m_p \frac{d^2 R}{d t^2} & =F=\frac{1}{3} \frac{N Y^2 e^2 R}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N R \\ & =\left(\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_p^2 N\right) R \\ \therefore\quad\frac{d^2 R}{d t^2} & =\frac{1}{m_p}\left[\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N\right] R \quad \text{... (iv)}\\ \text{where,}\quad\alpha^2 & =\frac{1}{m_p}\left[\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N\right] \end{aligned}$$
The general solution of Eq. (iv) is given by $R=A e^{\alpha t}+B e^{-\alpha t}$. We are looking for expansion, here, so $B=0$ and $R=A e^{\alpha t}$.
$\Rightarrow \quad$ Velocity of expansion, $v=\frac{d R}{d t}=A e^{\alpha t}(\alpha)=\alpha A e^{\alpha t}=\alpha R$
Hence, $v \propto$ R i.e., velocity of expansion is proportional to the distance from the centre.
Consider a sphere of radius $R$ with charge density distributed as $p(r)=k r$ for $r \leq R=0$ for $r>R$.
(a) Find the electric field as all points $r$.
(b) Suppose the total charge on the sphere is $2 e$ where $e$ is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
(a) Let us consider a sphere $S$ of radius $R$ and two hypothetic sphere of radius $r
Now, for point $r
$$\begin{aligned} & \oint E d S=\frac{1}{\varepsilon_0} \int \rho d V \\ & \text { [For } \left.d V, V=\frac{4}{3} \pi r^3 \Rightarrow d V=3 \times \frac{4}{3} \pi r^3 d r=4 \pi r^2 d r\right] \\ & \Rightarrow \quad \quad \quad \mathrm{E} . \mathrm{dS}=\frac{1}{\varepsilon_0} 4 \pi K \int_0^r r^3 d r \quad (\because p(r)=Kr)\\ & \Rightarrow \quad \text { (E) } 4 \pi r^2=\frac{4 \pi K}{\varepsilon_0} \frac{r^4}{4} \\ & \Rightarrow \quad E=\frac{1}{4 \varepsilon_0} K r^2 \end{aligned}$$
Here, charge density is positive.
So, direction of E is radially outwards.
For points $r>R$, electric field intensity will be given by
$$\begin{array}{rlrl} & & \rho E d S & =\frac{1}{\varepsilon_0} \int \rho \cdot d V \\ \Rightarrow \quad & E\left(4 \pi r^2\right) & =\frac{4 \pi K}{\varepsilon_0} \int_0^R r^3 d r=\frac{4 \pi K}{\varepsilon_0} \frac{R^4}{4} \\ \Rightarrow \quad & E & =\frac{K}{4 \varepsilon_0} \frac{R^4}{r^2} \end{array}$$
Charge density is again positive. So, the direction of E is radially outward.
(b) The two protons must be on the opposite sides of the centre along a diameter following the rule of symmetry. This can be shown by the figure given below. Charge on the sphere,
$$\begin{aligned} & q=\int_0^R \rho d V=\int_0^R(K r) 4 \pi r^2 d r \\ & q=4 \pi K \frac{R^4}{4}=2 e \\ \therefore\quad& K=\frac{2 e}{\pi R^4} \end{aligned}$$
If protons 1 and 2 are embedded at distance $r$ from the centre of the sphere as shown, then attractive force on proton 1 due to charge distribution is
$$F_1=e E=\frac{-e K r^2}{4 \varepsilon_0}$$
Repulsive force on proton 1 due to proton 2 is
$$\begin{aligned} F_2 & =\frac{e^2}{4 \pi \varepsilon_0(2 r)^2} \\ \text{Net force on proton 1,}\quad F & =F_1+F_2 \\ F & =\frac{-e K r^2}{4 \varepsilon_0}+\frac{e^2}{16 \pi \varepsilon_0 r^2} \\ \text{So,}\quad F & =\left[\frac{-e r^2}{4 \varepsilon_0} \frac{Z e}{\pi R^4}+\frac{e^2}{16 \pi \varepsilon_0 r^4}\right] \end{aligned}$$
Thus, net force on proton 1 will be zero, when
$$\begin{aligned} \frac{e r^2 2 e}{4 \varepsilon_0 \pi R^4} & =\frac{e^2}{16 \pi \varepsilon_0 r} \\ r^4 & =\frac{R^4}{8} \\ r & =\frac{R}{(8)^{1 / 4}} \end{aligned}$$
This is the distance of each of the two protons from the centre of the sphere.
Two fixed, identical conducting plates ( $\alpha$ and $\beta$ ), each of surface area $S$ are charged to $-Q$ and $q$, respectively, where $Q>q>0$. A third identical plate $(\gamma)$, free to move is located on the other side of the plate with charge $q$ at a distance $d$ (figure). The third plate is released and collides with the plate $\beta$. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst $\beta$ and $\gamma$.
(a) Find the electric field acting on the plate $\gamma$ before collision.
(b) Find the charges on $\beta$ and $\gamma$ after the collision.
(c) Find the velocity of the plate $\gamma$ after the collision and at a distance $d$ from the plate $\beta$.
(a) Net electric field at plate $\gamma$ before collision is equal to the sum of electric field at plate $\gamma$ due to plate $\alpha$ and $\beta$.
The electric field at plate $\gamma$ due to plate $\alpha$ is $E_1=\frac{-Q}{S\left(2 \varepsilon_0\right)}$, to the left.
The electric field at plate $\gamma$ due to plate $\beta$ is $E_2=\frac{q}{S\left(2 \varepsilon_0\right)}$, to the right.
Hence, the net electric field at plate $\gamma$ before collision.
$$E=E_1+E_2=\frac{q-Q}{S\left(2 \varepsilon_0\right)} \text {, to the left, if } Q>q$$
(b) During collision, plates $\beta$ and $\gamma$ are together. Their potentials become same.
Suppose charge on plate $\beta$ is $q_1$ and charge on plate $\gamma$ is $q_2$. At any point $O$, in between the two plates, the electric field must be zero.
Electric field at $O$ due to plate $\alpha=\frac{-Q}{S\left(2 \varepsilon_0\right)}$, to the left Electric field at $O$ due to plate $\beta=\frac{q_1}{S\left(2 \varepsilon_0\right)}$, to the right
Electric field at $O$ due to plate $\gamma=\frac{q_2}{S\left(2 \varepsilon_0\right)}$, to the left
As the electric field at $O$ is zero, therefore
$$\begin{aligned} \frac{Q+q_2}{S\left(2 \varepsilon_0\right)} & =\frac{q_1}{S\left(2 \varepsilon_0\right)} \\ \therefore\quad Q+q_2 & =q_1 \\ Q & =q_1-q_2\quad\text{... (i)} \end{aligned}$$
As there is no loss of charge on collision,
$$Q+q=q_1+q_2\quad\text{... (ii)}$$
$$\begin{aligned} &\text { On solving Eqs. (i) and (ii), we get }\\ &\begin{aligned} q_1 & =(Q+q / 2)=\text { charge on plate } \beta \\ q_2 & =(q / 2)=\text { charge on plate } \gamma \end{aligned} \end{aligned}$$
(c) After collision, at a distance $\alpha$ from plate $\beta$,
Let the velocity of plate $\gamma$ be $v$. After the collision, electric field at plate $\gamma$ is
$$E_2=\frac{-Q}{2 \varepsilon_0 S}+\frac{(Q+q / 2)}{2 \varepsilon_0 S}=\frac{q / 2}{2 \varepsilon_0 S} \text { to the right. }$$
Just before collision, electric field at plate $\gamma$ is $E_1=\frac{Q-q}{2 \varepsilon_0 S}$.
If $F_1$ is force on plate $\gamma$ before collision, then $F_1=E_1 Q=\frac{(Q-q) Q}{2 \varepsilon_0 S}$
Total work done by the electric field is round trip movement of plate $\gamma$
$$\begin{aligned} W & =\left(F_1+F_2\right) d \\ & =\frac{\left[(Q-q) Q+(q / 2)^2\right] d}{2 \varepsilon_0 S}=\frac{(Q-q / 2)^2 d}{2 \varepsilon_0 S} \end{aligned}$$
If $m$ is mass of plate $\gamma$, the KE gained by plate $\gamma=\frac{1}{2} m v^2$
According to work-energy principle, $\frac{1}{2} m v^2=W=\frac{(Q-q / 2)^2 d}{2 \varepsilon_0 S}$
$$\gamma=(Q-q / 2)\left(\frac{d}{m \varepsilon_0 S}\right)^{1 / 2}$$
There is another useful system of units, besides the SI/MKS. A system, called the CGS (Centimeter-Gram-Second) system. In this system, Coulomb's law is given by $\mathbf{F}=\frac{Q q}{r^2} \hat{\mathbf{r}}$. where the distance $r$ is measured in $\mathrm{cm}\left(=10^{-2} \propto\right), F$ in dynes $\left(=10^{-5} \mathrm{~N}\right)$ and the charges in electrostatic units (es units), where 1 es unit of charge $=\frac{1}{[3]} \times 10^{-9} \mathrm{C}$. The number [3] actually arises from the speed of light in vacuum which is now taken to be exactly given by $c=2.99792458 \times 10^8 \mathrm{~m} / \mathrm{s}$. An approximate value of $c$, then is $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$.
(i) Show that the Coulomb's law in CGS units yields 1 esu of charge $=1(\text { dyne })^{1 / 2} \mathrm{~cm}$. Obtain the dimensions of units of charge in terms of mass $M$, length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$.
(ii) Write 1 esu of charge $=x C$, where $x$ is a dimensionless number. Show that this gives $\frac{1}{4 \pi \varepsilon_0}=\frac{10^{-9}}{\mathrm{x}^2} \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$. With $x=\frac{1}{[3]} \times 10^{-9}$, we have $\frac{1}{4 \pi \varepsilon_0}=[3]^2 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}, \frac{1}{4 \pi \varepsilon_0}=(2.99792458)^2 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$ (exactly).
(i) From the relation, $F=\frac{Q q}{r^2}=1$ dyne $=\frac{[1 \text { esu of charge }]^2}{[1 \mathrm{~cm}]^2}$
So, 1 esu of charge $=(1 \text { dyne })^{1 / 2} \times 1 \mathrm{~cm}=F^{1 / 2} \cdot \mathrm{~L}=\left[\mathrm{MLT}^{-2}\right]^{1 / 2} \mathrm{~L}$
$\Rightarrow 1$ esu of charge $=M^{1 / 2} L^{3 / 2} T^{-1}$.
Thus, esu of charge is represented in terms of fractional powers $\frac{1}{2}$ of $M$ and $\frac{3}{2}$ of $L$.
(ii) Let 1 esu of charge $=x C$, where $x$ is a dimensionless number. Coulomb force on two charges, each of magnitude 1 esu separated by 1 cm is dyne $=10^{-5} \mathrm{~N}$. This situation is equivalent to two charges of magnitude $x \mathrm{C}$ separated by $10^{-2} \mathrm{~m}$.
$\therefore \quad F=\frac{1}{4 \pi \varepsilon_0} \frac{x^2}{\left(10^{-2}\right)^2}=1$ dyne $=10^{-5} \mathrm{~N}$
$\therefore \quad \frac{1}{4 \pi \varepsilon_0}=\frac{10^{-9}}{x^2} \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$
Taking, $\quad x=\frac{1}{|3| \times 10^9}$,
we get, $\quad \frac{1}{4 \pi \varepsilon_0}=10^{-9} \times|3|^2 \times 10^{18} \frac{\mathrm{Nm}^2}{\mathrm{C}^2}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$
If $|3| \rightarrow 2.99792458$, we get $\frac{1}{4 \pi \varepsilon_0}=8.98755 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}$.
Two charges $-q$ each are fixed separated by distance $2 d$. A third charge $q$ of mass $m$ placed at the mid-point is displaced slightly by $x(x< d)$ perpendicular to the line joining the two fixed charged as shown in figure. Show that $q$ will perform simple harmonic oscillation of time period.
$T=\left[\frac{8 \pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1 / 2}$
Let us elaborate the figure first.
Given, two charge $-q$ at $A$ and $B$
$$A B=A O+O B=2 d$$
$x=$ small distance perpendicular to $O$.
i.e., $x
Horizontal components of these forces $F_n$ are cancel out. Vertical components along $P O$ add.
If $\angle A P O=O$, the net force on $q$ along $P O$ is $F^{\prime}=2 F \cos Q$
$$\begin{aligned} & =\frac{2 q^2}{4 \pi \varepsilon_0 r^2}\left(\frac{x}{r}\right) \\ & =\frac{2 q^2 x}{4 \pi \varepsilon_0\left(d^2+x^2\right)^{3 / 2}} \\ & \text { When, } \\ & x<< d, F^{\prime}=\frac{2 q^2 x}{4 \pi \varepsilon_0 d^3}=K x \\ & \text { where, } \\ & K=\frac{2 q^2}{4 \pi \varepsilon_0 d^3} \\ & \Rightarrow \quad F \propto x \end{aligned}$$
i.e., force on charge $q$ is proportional to its displacement from the centre $O$ and it is directed towards $O$.
Hence, motion of charge $q$ would be simple harmonic, where
$$\begin{aligned} \omega & =\sqrt{\frac{K}{m}} \\ T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \\ & =2 \pi \sqrt{\frac{m \cdot 4 \pi \varepsilon_0 d^3}{2 q^2}}=\left[\frac{8 \pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1 / 2} \end{aligned}$$