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
$$\begin{aligned} \omega & =\sqrt{\frac{K}{m}} \\ T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \\ & =2 \pi \sqrt{\frac{m \cdot 4 \pi \varepsilon_0 d^3}{2 q^2}}=\left[\frac{8 \pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1 / 2} \end{aligned}$$
Total charge $-Q$ is uniformly spread along length of a ring of radius $R$.A small test charge $+q$ of mass $m$ is kept at the centre of the ring and is given a gentle push along the axis of the ring.
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.
Let us draw the figure according to question,
A gentle push on q along the axis of the ring gives rise to the situation shown in the figure below.
Taking line elements of charge at $A$ and $B$, having unit length, then charge on each elements.
$$d F=2\left(-\frac{Q}{2 \pi R}\right) q \times \frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cos \theta$$
Total force on the charge $q$, due to entire ring
$$\begin{aligned} & F=-\frac{Q q}{\pi R}(\pi R) \cdot \frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cdot \frac{2}{r} \\ \text{Here}, Z<< R, \quad & F=-\frac{Q q z}{4 \pi \varepsilon_0\left(Z^2+R^2\right)^{3 / 2}} \end{aligned}$$
where
$$\begin{gathered} \frac{Q q}{4 \pi \varepsilon_0 R^3}=\text { constant } \\ \Rightarrow\quad F \propto-Z \end{gathered}$$
Clearly, force on $q$ is proportional to negative of its displacement. Therefore, motion of $q$ is simple harmonic.
$$\begin{aligned} & \omega=\sqrt{\frac{K}{m}} \text { and } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \\ & T=2 \pi \sqrt{\frac{m 4 \pi \varepsilon_0 R^3}{Q q}} \\ \Rightarrow\quad & T=2 \pi \sqrt{\frac{4 \pi \varepsilon_0 m R^3}{Q q}} \end{aligned}$$