Electrostatics

As, $R$ be the radius of the ring. Consider a small strip of length $d l$ having charge $d q$ lying at an angle $\theta$.

$d l=R d \theta$

Charge on $d l=\lambda R d \theta$

Force at $1 \mathrm{~C}$ due to $\mathrm{dl}$

$=\frac{k \lambda R d \theta}{R^2}=\frac{k \lambda}{R} d \theta=d F$

We need to consider only the component $d F \cos \theta$, as the component $d F \sin \theta$ will cancel out because of the symmetrical element $d l$.

The total force on $1 \mathrm{C}$ is

$\begin{aligned} F & =\int_\limits{-\pi / 2}^{\pi / 2} d F \cos \theta \\ & =\frac{k \lambda}{R} \int_\limits{-\pi / 2}^{\pi / 2} \cos \theta d \theta \\ & =\frac{k \lambda}{R} \times 2=\frac{2 k \lambda}{R} \end{aligned}$

When the negative charge is shifted at a distance $x$ from the centre of the ring along its axis, then force acting on the point charge due to the ring.

$\begin{aligned} F & =q E ~(\text {towards centre}) \\\\ & =q \cdot \frac{k Q x}{\left(R^2+x^2\right)^{3 / 2}} \end{aligned}$

If $R >> x$, then $R^2+x^2 \simeq R^2$

and $\quad F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q x}{R^3}$ (towards centre)

$\Rightarrow \quad a=\frac{F}{m}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q x}{m R^3}$

Since, restoring force $F_E \propto x$, therefore motion of charge particle will be SHM.

Time period of SHM,

$T=\frac{2 \pi}{\omega}=\left[\frac{16 \pi^3 \varepsilon_0 R^3 m}{Q q}\right]^{1 / 2}\left[\because \omega^2=\frac{Q q}{4 \pi \varepsilon_0 R^3}\right]$

Negatively charged body contains more electron than its natural state of mass is slightly increased.

In Ist case, when charge +Q is situated at C

Electric potential energy of system

$U_1=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q) Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}$

In Ind case, when charge $+Q$ is moved from $C$ to $D$

Electric potential energy of system in this case

$U_2=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q)(Q)}{L}$

Work done $(\Delta U)=U_2-U_1$

$\begin{aligned} = & \left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}\right) \\ & -\left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}\right) \\ = & \frac{q Q}{4 \pi \varepsilon_0}\left[\frac{1}{3 L}-\frac{1}{L}\right]=\frac{q Q}{4 \pi \varepsilon_0} \frac{(1-3)}{3 L} \\ = & -\frac{2 q Q}{12 \pi \varepsilon_0 L}=-\frac{q Q}{6 \pi \varepsilon_0 L} \end{aligned}$

We know that, for an electric dipole

$\begin{aligned} E_{\text {axial }} & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 p}{r^3}\right) \quad \text{..... (i)}\\ \text{and} \quad E_{\text {equatorial }} & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{p}{r^3}\right) \quad \text{.... (ii)} \end{aligned}$

Hence, from Eqs. (i) and (ii), we get

$\frac{E_{\text {axil }}}{2}=E_{\text {equatorial }}$

Hence, $\quad E_{\text {equatorial }}=E / 2$

Reason is false as electric field due to dipole varies inversely as cube of distance, i.e.

$E \propto \frac{1}{r^3}$