(i) From the given figure, we can analyse that the chlorine atom is at the centre of the cube i.e., at equal distance from all the eight corners of cube where cesium atoms are placed. Thus, due to symmetry the forces due to all Cs tons, on Cl atom will cancel out.

Hence,

$$\begin{aligned} & E=\frac{F}{q} \text { where } F=0 \\ \therefore\quad & E=0 \end{aligned}$$

(ii) Thus, net force on Cl atom at $A$ would be,

$$F=\frac{e^2}{4 \pi \varepsilon_0 r^2}$$

where, $r=$ distance between Cl ion and Cs ion.

Applying Pythagorous theorem, we get

$$\begin{aligned} r & =\sqrt{(0.20)^2+(0.20)^2+(0.20)^2} \times 10^{-9} \mathrm{~m} \\ & =0.346 \times 10^{-9} \mathrm{~m} \end{aligned}$$

Now,

$$\begin{aligned} F & =\frac{q^2}{4 \pi \varepsilon_0 r^2}=\frac{e^2}{4 \pi \varepsilon_0 r_2} \\ & =\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{\left(0.346 \times 10^{-9}\right)^2}=1.92 \times 10^{-9} \mathrm{~N} \end{aligned}$$

Here, let us keep the change 2q at a distance r from A.

Thus, charge $2 q$ will not experience any force.

When, force of repulsion on it due to $q$ is balanced by force of attraction on it due to $-3 q$, at $B$, where $A B=d$.

Thus, force of attraction by $-3 q=$ Force of repulsion by $q$

$$\begin{aligned} \Rightarrow \quad& \frac{2 q \times q}{4 \pi \varepsilon_0 x^2} =\frac{2 q \times 3 q}{4 \pi \varepsilon_0(x+d)^2} \\ \Rightarrow \quad& (x+d)^2 =3 x^2 \\ \Rightarrow \quad& x^2+d^2+2 x d =3 x^2 \\ \Rightarrow \quad& =2 x^2-d^2 \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \therefore \quad 2 x^2-2 d x-d^2 & =0 \\ x & =\frac{d}{2} \pm \frac{\sqrt{3} d}{2} \end{aligned}\\ &\text { (Negative sign be between } q \text { and }-3 q \text { and hence is unadaptable.) }\\ &\begin{aligned} x & =-\frac{d}{2}+\frac{\sqrt{3} d}{2} \\ & =\frac{d}{2}(1+\sqrt{3}) \text { to the left of } q . \end{aligned} \end{aligned}$$

(i) Here, in the figure, the electric lines of force emanate from $A$ and $C$. Therefore, charges $A$ and $C$ must be positive.

(ii) The number of electric lines of forces enamating is maximum for charge $C$ here, so $C$ must have the largest magnitude.

(iii) Point between two like charges where electrostatic force is zero is called netural point. So, the neutral point lies between $A$ and $C$ only.

Now the position of neutral point depends on the strength of the forces of charges. Here, more number of electric lines of forces shows higher strength of charge $C$ than $A$. So, neutral point lies near $A$.

(a) (i) The point $O$ is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry, the forces due to all the charges are cancelled out. As a result electric field at $O$ is zero.

(ii) When charge $q$ is removed a negative charge will develop at $A$ giving electric field $E=\frac{q \times 1}{4 \pi \varepsilon_0 r^2}$ along $O A$.

(iii) If charge $q$ at $A$ is replaced by $-q$, then two negative charges $-2 q$ will develop there. Thus, the value of electric field $E=\frac{2 q}{4 \pi \varepsilon_0 r^2}$ along $O A$.

(b) When pentagon is replaced by $n$ sided regular polygon with charge $q$ at each of its corners, the electric field at $O$ would continue to be zero as symmetricity of the charges is due to the regularity of the polygon. It doesn't depend on the number of sides or the number of charges.

(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.