Physics 5 Marks Questions

Electric field at any point on the axis at a distance x from the center of the ring is

$\text{E}=\frac{\text{xQ}}{4\pi\in_0\big(\text{R}^2+\text{x}^2\big)^{\frac{3}{2}}}$

$=\frac{\text{KxQ}}{\big(\text{R}^2+\text{x}^2\big)^{\frac{3}{2}}}$

Differentiating with respect to x

$\frac{\text{dE}}{\text{dx}}=\frac{\text{KQ}\big(\text{R}^2+\text{x}^2\big)-\text{KxQ}\big(\frac{3}{2}\big)\big(\text{R}^2+\text{x}^2\big)^{\frac{11}{2}}2\text{x}}{\big(\text{r}^2+\text{x}^2\big)^3}$

Since at a distance x, Electric field is maximum.

$\frac{\text{dE}}{\text{dx}}=0$

$\Rightarrow\text{KQ}\big(\text{R}^2+\text{x}^2\big)^{\frac{3}{2}}-\text{Kx}^2\text{Q}^3\big(\text{R}^2+\text{x}^2\big)^{\frac{1}{2}}=0$

$\Rightarrow\text{KQ}\big(\text{R}^2+\text{x}^2\big)^{\frac{3}{2}}=\text{Kx}^2\text{Q}3\big(\text{R}^2+\text{x}^2\big)^{\frac{1}{2}}$

$\Rightarrow\text{R}^2+\text{x}^2=3\text{x}^2$

$\Rightarrow2\text{x}^2=\text{R}^2$

$\Rightarrow\text{x}^2=\frac{\text{R}^2}{2}$

$\Rightarrow\text{x}=\frac{\text{R}}{\sqrt{2}}$

1. $\text{p}=2\text{qa}$

2. $\text{E}_1\sin\theta=\text{E}_2\sin\theta$ Electric field intensity

$=\text{E}_1=\frac{\text{Kqp}}{\text{a}^2+\text{d}^2}$

So, $\text{E}=\frac{2\text{KPQ}}{\text{a}^2+\text{d}^2}\frac{\text{a}}{\big(\text{a}^2+\text{d}^2\big)^{\frac{1}{2}}}$

$=\frac{2\text{Kq}\times\text{a}}{\big(\text{a}^2+\text{d}^2\big)^{\frac{3}{2}}}$

When a << d

$=\frac{2\text{Kqa}}{\big(\text{d}^2\big)^{\frac{3}{2}}}$

$=\frac{\text{PK}}{\text{d}^3}$

$=\frac{1}{4\pi\in_0}\frac{\text{p}}{\text{d}^3}$

Radius = r

So, $2\pi\text{r}=$ Circumference

Charge density $=\lambda$

Total charge $=2\pi\text{r}\times\lambda$

Electric potential $=\frac{\text{Kq}}{\text{r}}$

$=\frac{1}{4\pi\in_0}\times\frac{2\pi\text{r}\lambda}{\big(\text{x}^2+\text{r}2\big)^{\frac{1}{2}}}$

$=\frac{\text{r}\lambda}{2\in_0\big(\text{x}^2+\text{r}^2\big)^{\frac{1}{2}}}$

So, Electric field $=\frac{\text{V}}{\text{r}}\cos\theta$

$=\frac{\text{r}\lambda}{2\in_0\big(\text{x}^2+\text{r}^2\big)^{\frac{1}{2}}}\times\frac{\text{x}}{\big(\text{x}^2+\text{r}^2\big)^{\frac{1}{2}}}$

$=\frac{\text{r}\lambda\text{x}}{2\in_0\big(\text{x}^2+\text{r}^2\big)^{\frac{3}{2}}}$

$\text{q}_\text{A}=2\times10^{-6}\text{C},\ \text{M}_\text{b}=80\text{g},\ \mu=0.2$

Since B is at equilibrium,

So, $\text{Fe}=\mu\text{R}$

$\Rightarrow\frac{\text{Kq}_\text{A}\text{q}_\text{B}}{\text{r}^2}=\mu\text{R}=\mu\text{m}\times\text{g}$

$\Rightarrow\frac{9\times10^9\times2\times10^{-6}\times\text{q}_\text{B}}{0.01}=0.2\times0.08\times9.8$

$\Rightarrow\text{q}_\text{B}=\frac{0.2\times0.08\times9.8\times0.01}{9\times10^9\times2\times10^{-6}}$

$=8.7\times10^{-8}\text{C}$

Range $=\pm8.7\times10^{-8}\text{C}$

$\text{m}=40\text{g},\ \text{q}=4\times10^{-6}\text{C}$

Time for 20 oscillations $=45\sec.$ Time for 1 oscillation $=\frac{45}{20}\sec$

When no electric field is applied, $\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$

$\Rightarrow\frac{45}{20}=2\pi\sqrt{\frac{\ell}{10}}$

$\Rightarrow\frac{\ell}{10}=\Big(\frac{45}{20}\Big)^2\times\frac{1}{4\pi^2}$

$\Rightarrow\ell=\frac{(45)^2\times10}{(20)^2\times4\pi^2}$

$=1.2836$

Time for 1 oscillation $= 2.598$

Time for 20 oscillation $=2.598\times20$

$=51.96\sec$

$=52\sec.$

$\text{q}=1.0\times10^{-8}\text{C},\ \ell=20\text{cm}$

$\text{E}=?,\ \text{V}=?$

Since it forms an equipotential surface.

So the electric field at the centre is Zero.

$\text{r}=\frac{2}{3}\sqrt{\big(2\times10^{-1}\big)^2-\big(10^{-1}\big)^2}$

$=\frac{2}{3}\sqrt{4\times10^{-2}-10^{-2}}$

$=\frac{2}{3}\sqrt{10^{-2}(4-1)}$

$=\frac{2}{3}\times10^{-2}\times1.732$

$=1.15\times10^{-1}$

$\text{V}=\frac{3\times9\times10^9\times1\times10^{-8}}{1\times10^{-1}}$

$=23\times10^2$

$2.3\times10^3\text{V}$

We know: Electric field ‘E’ at ‘P’ due to the charged ring

$=\frac{\text{KQx}}{\big(\text{R}^2+\text{x}^2\big)^{\frac{3}{2}}}$

$=\frac{\text{KQx}}{\text{R}^3}$

Force experienced ‘F’ $=\text{Q}\times\text{E}=\frac{\text{q}\times\text{K}\times\text{Qx}}{\text{R}^3}$

Now, amplitude = x

So, $\text{T}=2\pi\sqrt{\frac{\text{x}}{\frac{\text{KQqx}}{\text{mR}^3}}}$

$=2\pi\sqrt{\frac{\text{mR}^3\text{x}}{\text{KQqx}}}$

$=2\pi\sqrt{\frac{4\pi\in_0\text{mR}^3}{\text{Qq}}}$

$=\sqrt{\frac{4\pi^2\times4\pi\in_0\text{mR}^3}{\text{qQ}}}$

$\Rightarrow\text{T}=\bigg[\frac{16\pi^3\in_0\text{mR}^3}{\text{qQ}}\bigg]^{\frac{1}{2}}$

Given,

u = Velocity of projection, $\vec{\text{E}}$ = Electric field intensity

q = Charge; m = mass of particle

We know, Force experienced by a particle with charge ‘q’ in an electric field $\vec{\text{E}}=\text{qE}$

$\therefore\ $acceleration produced $=\frac{\text{qE}}{\text{m}}$

As the particle is projected against the electric field, hence deceleration $=\frac{\text{qE}}{\text{m}}$

So, let the distance covered be ‘s'

Then, v² = u² + 2as [where a = acceleration, v = final velocity]

Here $0=\text{u}^2-2\times\frac{\text{qE}}{\text{m}}\times\text{S}$

$\Rightarrow\text{S}=\frac{\text{u}^2\text{m}}{2\text{qE}}\text{units}$

$\lambda=$ Charge per unit length $=\frac{\text{Q}}{\text{L}}$

dq₁ for a length d $=\lambda\times\text{dl}$

Electric field at the centre due to charge $=\text{k}\times\frac{\text{dq}}{\text{r}^2}$

The horizontal Components of the Electric field balances each other. Only the vertical components remain.

$\therefore\ $Net Electric field along vertical

$\text{d}_\text{E}=2\text{E}\cos\theta$

$=\frac{\text{Kdq}\times\cos\theta}{\text{r}^2}$

$=\frac{2\text{k}\cos\theta}{\text{r}^2}\times\lambda\times\text{dl}$ $\Big[$but $\text{d}\theta=\frac{\text{d}\ell}{\text{r}}=\text{d}\ell=\text{rd}\theta\Big]$

$\Rightarrow\frac{2\text{k}\lambda}{\text{r}^2}\cos\theta\times\text{rd}\theta=\frac{2\text{k}\lambda}{\text{r}}\cos\theta\times\text{d}\theta$

$\text{E}=\int\limits^{\frac{\pi}{2}}_0\frac{2\text{k}\lambda}{\text{r}}\cos\theta\times\text{d}\theta$

$=\int\limits_0^{\frac{\pi}{2}}\frac{2\text{k}\lambda}{\text{r}}\sin\theta$

$=\frac{2\text{k}\lambda\text{l}}{\text{r}}$

$=\frac{2\text{K}\theta}{\text{Lr}}$

but $\text{L}=\pi\text{R}\Rightarrow\text{r}=\frac{\text{L}}{\pi}$

So, $\text{E}=\frac{2\text{k}\theta}{\text{L}\times\big(\frac{\text{L}}{\pi}\big)}$

$=\frac{2\text{k}\pi\theta}{\text{L}^2}$

$=\frac{2}{4\pi\in_0}\times\frac{\pi\theta}{\text{L}^2}$

$=\frac{\theta}{2\in_0\text{L}^2}$

1. Dipolemoment $=\text{q}\times\ell$

(Where q = magnitude of charge $\ell=$ Separation between the charges)

$=2\times10^{-6}\times10^{-2}\text{cm}$

$=2\times10^{-8}\text{cm}$

2. We know, Electric field at an axial point of the dipole

$=\frac{2\text{KP}}{\text{r}^3}$

$=\frac{2\times9\times10^9\times2\times10^{-8}}{\big(1\times10^{-2}\big)^3}$

$=36\times10^7\text{N/C}$

3. We know, Electric field at a point on the perpendicular bisector about 1m away from centre of dipole.

$=\frac{\text{KP}}{\text{r}^3}$

$=\frac{9\times10^9\times2\times10^{-8}}{1^3}$

$=180\text{N/C}$

When the charge is placed at A,

$\text{E}_1=\frac{\text{Kq}_1\text{q}_2}{\text{r}}+\frac{\text{Kq}_3\text{q}_4}{\text{r}}$

$=\frac{9\times10^9(2\times10^{-7})^2}{0.1}+\frac{9\times10^9(2\times10^{-7})^2}{0.1}$

$=\frac{2\times9\times10^9\times4\times10^{-14}}{0.1}$

$=72\times10^{-4}\text{J}$

When charge is placed at B,

$\text{E}_2=\frac{\text{Kq}_1\text{q}_2}{\text{r}}+\frac{\text{Kq}_3\text{q}_4}{\text{r}}$

$=\frac{2\times9\times10^9\times4\times10^{-14}}{0.2}$

$=36\times10^{-4}\text{J}$

1. The Electric field is along x-direction

Thus potential difference between,

$\text{A}=(0, 0);\ \text{B}=(4, 2)$

$\Rightarrow\delta\text{V}=-\text{E}\times\delta\text{x}$

$=-20\times(40)$

$=-80\text{V}$

Potential energy (U_B - U_A) between the points $=\delta\text{V}\times\text{q}$

$=-80\times(-2)\times10^{-4}$

$=160\times10^{-4}$

$=0.016\text{J}$

2. $\text{A}=(4\text{m},2\text{m});\ \text{B}=(6\text{m},5\text{m})$

$\Rightarrow\delta\text{V}=-\text{E}\times\delta\text{x}$

$=-20\times2$

$=-40\text{V}$

Potential energy (U_B - U_A) between the points $=\delta\text{V}\times\text{q}$

$=-40\times(-2)\times10^{-4})$

$=80\times10^{-4}$

$=0.008\text{J}$

3. $\text{A}=(0,0);\ \text{B}=(6\text{m}, 5\text{m})$

$\Rightarrow\delta\text{V}=-\text{E}\times\delta\text{x}$

$=-20\times6$

$=-120\text{V}$

Potential energy (U_B - U_A) between the points A and B

$\Rightarrow\delta\text{V}\times\text{q}$

$=-120\times(-2\times10^{-4})$

$=240\times10^{-4}$

$=0.024\text{J}$

q¹ = q² = q³ = q⁴ = 2 × 10^-6C

$\text{v}=5\text{cm}=5\times10^{-2}\text{m}$

So force on $\bar{\text{C}}=\bar{\text{F}}_\text{CA}+\bar{\text{F}}_\text{CB}+\bar{\text{F}}_\text{CD}$

So Force along × Component $=\bar{\text{F}}_\text{CD}+\bar{\text{F}}_\text{CA}\cos45^\circ+0$

$=\frac{\text{k}\big(2\times10^{-6}\big)^2}{\big(5\times10^{-2}\big)^2}+\frac{\text{k}\big(2\times10^{-6}\big)^2}{\big(5\times10^{-2}\big)^2}\frac{1}{2\sqrt{2}}$

$=\text{kq}^2\Big(\frac{1}{25\times10^{-4}}+\frac{1}{50\sqrt{2}\times10^{-4}}\Big)$

$\frac{9\times10^9\times4\times10^{12}}{24\times10^{-4}}\Big(1+\frac{1}{2\sqrt{2}}\Big)$

$=1.44(1.35)=19.49$ Force along % component = 19.49

So, Resultant $\text{R}=\sqrt{\text{Fx}^2+\text{Fy}^2}$

$=19.49\sqrt{2}$

$=27.56$

Let the charge on C = q

So, net force on c is equal to zero

So, $\text{F}_{\overline{\text{AC}}}+\text{F}_{\overline{\text{BA}}}=0,$

But $\text{F}_{\overline{\text{AC}}}=\text{F}_{\overline{\text{BC}}}$

$\Rightarrow\frac{\text{kqQ}}{\text{x}^2}=\frac{\text{k}2\text{qQ}}{(\text{d}-\text{x})^2}$

$\Rightarrow2\text{x}^2=(\text{d}-\text{x})^2$

$\Rightarrow\sqrt{2}\text{x}=\text{d}-\text{x}$

$\Rightarrow\text{x}=\frac{\text{d}}{\sqrt{2}+1}$

$=\frac{\text{d}}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}$

$=\text{d}\big(\sqrt{2}-1\big)$

For the charge on rest, $\text{F}_\text{AC}+\text{F}_\text{AB}=0$

$(2.414)^2\frac{\text{kqQ}}{\text{d}^2}+\frac{\text{kq}(2\text{q})}{\text{d}^2}=0$

$\Rightarrow\frac{\text{kq}}{\text{d}^2}\big[(2.414)^2\text{Q}+2\text{q}\big]=0$

$\Rightarrow2\text{q}=-(2.414)^2\text{Q}$

$\Rightarrow\text{Q}=\frac{2}{-\big(\sqrt{2}+1\big)^2}\text{q}$

$=-\Big(\frac{2}{3+2\sqrt{2}}\Big)\text{q}$

$=-(0.343)\text{q}$

$=-\big(6-4\sqrt{2}\big)$

1. $\text{F}=\text{qE}=2.5\times10^{-4}\times1.2\times10^4=3\text{N}$

So, $\text{a}=\frac{\text{F}}{\text{m}}$

$=\frac{3}{10^{-3}}$

$=3\times10^3$

$\text{E}_\text{q}=\text{mg}=10^{-3}\times9.8\times10^{-3}\text{N}$

2. $\text{S}=\frac{1}{2}\text{at}^2$

$\text{t}=\sqrt{\frac{2\text{a}}{9}}$

$=\sqrt{\frac{2\times4\times10^{-1}}{3\times10^{3}}}$

$=1.63\times10^{-2}\text{sec}$

3. $\text{v}^2=\text{u}^2+2\text{as}$

$=0+2\times3\times10^3\times4\times10^{-1}$

$=24\times10^2$

$\Rightarrow\text{v}=\sqrt{24\times10^2}$

$4.9\times10=49\text{m/sec}$

4. Work done by the electric force $\text{w}=\text{F}\rightarrow\text{td}$

$=3\times4\times10^{-1}$

$=12\times10^{-1}$

$=1.2\text{J}$

1. $\text{A}=\frac{\text{Volt}}{\text{m}^2}$

$=\frac{\text{ML}^2\text{T}^2}{\text{ITL}^2}$

$\big[\text{MT}^{-3}\text{I}^{-1}\big]$

2. $\text{E}=\frac{\delta\text{V}\hat{\text{i}}}{\delta\text{x}}-\frac{\delta\text{V}\hat{\text{j}}}{\delta\text{y}}-\frac{\delta\text{V}\hat{\text{k}}}{\delta\text{z}}$

$=-\Big[\frac{\delta}{\delta\text{y}}\text{A}(\text{xy}+\text{yz}+\text{zx})+\frac{\delta}{\delta\text{y}}\text{A}(\text{xy}+\text{yz}+\text{zx})+\frac{\delta}{\delta\text{z}}\text{A}(\text{xy}+\text{yz}+\text{zx})\Big]$

$=-\Big[(\text{ay}+\text{Az})\hat{\text{i}}+(\text{Ax}+\text{Az})\hat{\text{i}}+(\text{Ax}+\text{Az})\hat{\text{j}}+(\text{Ay}+\text{Ax})\hat{\text{k}}\Big]$

$=-\text{A}(\text{y}+\text{z})\hat{\text{i}}+\text{A}(\text{x}+\text{z})\hat{\text{j}}+\text{A}(\text{y}+\text{x})\hat{\text{k}}$

3. $\text{A}=10\text{SI unit},\ \text{r}=(1\text{m},1\text{m},1\text{m})$

$\text{E}=-10(2)\hat{\text{i}}-10(2)\hat{\text{j}}-10(2)\hat{\text{ik}}$

$=\sqrt{20^2+20^2+20^2}$

$=\sqrt{1200}$

$=34.64\approx35\text{N/C}$

1. $\text{V}=\text{E}\times\text{d}\ell$

$=1000\times\frac{2}{100}$

$=20\text{V}$

2. $\text{u}=?,\ \vec{\text{E}}=1000,$

$=\frac{2}{100}\text{m}$

$\text{a}=\frac{\text{F}}{\text{m}}$

$=\frac{\text{q}\times\text{E}}{\text{m}}$

$=\frac{1.6\times10^{-19}\times1000}{9.1\times10^{-31}}$

$=1.75\times10^{14}\text{m/s}^2$

$0=\text{u}^2-2\times1.75\times10^{14}\times0.02$

$\Rightarrow\text{u}^2=0.04\times1.75\times10^{14}$

$\Rightarrow\text{u}=2.64\times10^6\text{m/s}.$

3. Now, $\text{U}=\text{u}\cos60^\circ,\ \text{V}=0,\ \text{s}=?$

$\text{a}=1.75\times10^{14}\text{m/s}^2,\ \text{V}^2=\text{u}^2-2\text{as}$

$\Rightarrow\text{s}=\frac{(\text{u}\cos60^\circ)}{2\times\text{a}}$

$=\frac{\Big(2.64\times10^6\times\frac{1}{2}\Big)^2}{2\times1.75\times10^{14}}$

$=\frac{1.75\times10^{12}}{3.5\times10^{14}}$

$=0.497\times10^{-2}\approx0.005\text{m}\approx0.50\text{cm}$

1. $\text{R}=\sqrt{\text{F}_\text{g}^2+\text{F}_\text{e}^2}$

$\sqrt{(0.1\times9.8)^2+(4.9\times10^{-5}\times2\times10^4)^2}$

$=\sqrt{0.9604+96.04\times10^{-2}}$

$=\sqrt{1.9208}$

$=1.3859\text{N}$

2. $\tan\theta=\frac{\text{F}_\text{g}}{\text{F}_\text{e}}$

$=\frac{\text{mg}}{\text{qE}}$

$=1$

So, $\theta=45^\circ$

$\therefore\ $Hence path is straight along resultant force at an angle 45^° with horizontal

Disp. Vertical $=\Big(\frac{1}{2}\Big)\times9.8\times2\times2$

$=19.6\text{m}$

3. Disp. Horizontal = $\text{S}=\Big(\frac{1}{2}\Big)\text{at}^2$

$=\frac{1}{2}\times\frac{\text{qE}}{\text{m}}\times\text{t}^2$

$=\frac{1}{2}\times\frac{0.98}{0.1}\times2\times2$

$=19.6\text{m}$

Net Dispt. $=\sqrt{(19.6)^2+(19.6)^2}$

$=\sqrt{768.32}$

$=27.7\text{m}$

1. Potential aat the origin is O.

$\text{dV}=-\text{E}_\text{x}\ \text{dx}-\text{E}_\text{y}\ \text{dy}-\text{E}_\text{z}\text{dz}$

$\Rightarrow\text{V}-0=-2\text{x}$

$\Rightarrow\text{V}=-2\text{x}$

2. $(25-0)=-2\text{x}$

$\Rightarrow\text{x}=-12.5\text{m}$

3. If potential at origin is 100 v, 

$\Rightarrow\text{V}-100=-2\text{x}$

$\Rightarrow\text{V}=-2\text{x}+100$

$=100-2\text{x}$

4. Potential at $\infty$ is 0,

$\Rightarrow\text{V}'=-2\text{x}$

$\Rightarrow\text{V}'=\text{V}+2\text{x}$

$=0+2\infty$

$\Rightarrow\text{V}'=\infty$

Potential at origin is $\infty.$ No, it is not practical to take potential at $\infty$ to be zero.

1. The angle between potential $\text{E}=\text{d}\ell=\text{dv}$

Change in potential = $10\text{V}=\text{dV}$

As $\text{E}=\bot\text{r}\text{dV}$ (As potential surface)

So, $\text{E}(10\times10^{-2})$

$\Rightarrow\text{E}\text{d}\ell\cos(90^\circ+30^\circ)=-\text{dv}$

$\Rightarrow\text{E}=\frac{-\text{dV}}{10\times10^{-2}\cos120^\circ}$

$=-\frac{10}{10^{-1}\times\Big(\frac{-1}{2}\Big)}$

$=200\text{V/m}$ making an angle 120^° with y-axis

2. As Electric field intensity is $\bot\text{r}$ to Potential surface

So, $\text{E}=\frac{\text{kq}}{\text{r}^2}\text{r}=\frac{\text{kq}}{\text{r}}$

$\Rightarrow\frac{\text{kq}}{\text{r}}=60\text{v}$

$\text{q}=\frac{6}{\text{K}}$

So, $\text{E}=\frac{\text{kq}}{\text{r}^2}$

$=\frac{6\times\text{k}}{\text{k}\times\text{r}^2}\text{v.m}$

$=\frac{6}{\text{r}^2}\text{v.m}$

Electric force $=\frac{\text{kq}^2}{(\ell\sin\text{Q}+\ell\sin\text{Q})^2}$

$=\frac{\text{kq}^2}{4\ell^2\sin^2}$

So, $\text{T}\cos\theta=\text{ms}$ (For equilibrium) $\text{T}\sin\theta=\text{Ef},$

$\text{tan}\theta=\frac{\text{Ef}}{\text{mg}}$

$\Rightarrow\text{mg}=\text{Ef}\cot\theta$

$=\frac{\text{kq}^2}{4\ell^2\sin^2\theta}\cot\theta$

$=\frac{\text{q}^2\cot\theta}{\ell^2\sin^2\theta16\pi\text{E}_0}$

$\text{m}=\frac{\text{q}^2\cot\theta}{16\pi\text{E}_0\ell^2\sin^2\theta\text{g}}\text{unit}$

$\text{F}_\text{AC}=\frac{\text{KQq}}{(\ell+\text{x})^2}$

$\text{F}_\text{CA}=\frac{\text{KQq}}{(\ell-\text{x})^2}$

Net force $=\text{KQq}\Big[\frac{1}{(\ell-\text{x})^2}-\frac{1}{(\ell-\text{x})^2}\Big]$

$=\text{KQq}\bigg[\frac{(\ell+\text{x})^2-(\ell-\text{x})^2}{(\ell+\text{x})^2(\ell-\text{x})^2}\bigg]$

$=\text{KQq}\bigg[\frac{4\ell\text{x}}{(\ell^2-\text{x}^2)^2}\bigg]$

$\text{x}<<<\text{l}=\frac{\text{d}}{2}$ neglecting x w.r.t. $\ell$ We get

net $\text{F}=\frac{\text{KQq}4\ell\text{x}}{\ell^4}=\frac{\text{KQq}4\text{x}}{\ell^3}$ acceleration $=\frac{4\text{KQqx}}{\text{m}\ell^3}$

Time period $=2\pi\sqrt{\frac{\text{displacement}}{\text{acceleration}}}$

$=2\pi\sqrt{\frac{\text{xm}\ell^3}{4\text{KQqx}}}$

$=2\pi\sqrt{\frac{\text{m}\ell^3}{4\text{KQq}}}$

$=\sqrt{\frac{4\pi^2\text{m}\ell^34\pi\in_0}{4\text{Qq}}}$

$=\sqrt{\frac{4\pi^3\text{m}\ell^3\in_0}{\text{Qq}}}$

$=\sqrt{4\pi^3\text{md}^3\in_08\text{Qq}}$

$=\bigg[\frac{\pi^3\text{md}^3\in_0}{2\text{Qq}}\bigg]^{\frac{1}{2}}$

$\text{q}=2.0\times10^{-8}\text{c},\ \text{n}=?,\ \sin\theta=\frac{1}{20}$

Force between the charges

$\text{F}=\frac{\text{Kq}_1\text{q}_2}{\text{r}^2}$

$=\frac{9\times10^9\times2\times10^{-8}\times2\times10^{-8}}{\big(3\times10^{-2}\big)^2}$

$=4\times10^{-3}\text{N}$

$\text{mg}\sin\theta=\text{F}$

$\Rightarrow\text{m}=\frac{\text{F}}{\text{g}\sin\theta}$

$=\frac{4\times10^{-3}}{10\times\Big(\frac{1}{20}\Big)}$

$=8\times10^{-3}$

$=8\text{gm}$

$\cos\theta=\sqrt{1-\sin^2\theta}$

$=\sqrt{1-\frac{1}{400}}$

$=\sqrt{\frac{400-1}{400}}$

$=0.99=1$

So, $\text{T}=\text{mg}\cos\theta$

$\text{T}=8\times10^{-3}10\times0.99$

$=8\times10^{-2}\text{M}$

$\text{q}_1=2\times10^{-6}\text{c}$ Let the distance be r unit

$\therefore\text{F}_\text{repulsion}=\frac{\text{kq}_1\text{q}_2}{\text{r}^2}$

For equilibrium $\frac{\text{kq}_1\text{q}_2}{\text{r}^2}=\text{mg}\sin\theta$

$\Rightarrow\frac{9\times10^9\times4\times10^{-12}}{\text{r}^2}=\text{m}\times9.8\times\frac{1}{2}$

$\Rightarrow\text{r}^2=\frac{18\times4\times10^{-3}}{\text{m}\times9.8}$

$=\frac{72\times10^{-3}}{9.8\times10^{-1}}$

$=7.34\times10^{-2}\ \text{metre}$

$\Rightarrow\text{r}=2.70924\times10^{-1}\text{metre}$ from the bottom

$\text{T}\cos\theta=\text{mg}\ \dots(1)$

$\text{T}\sin\theta=\text{Fe}\ \dots(2)$

Solving, $\frac{(2)}{(1)}$ we get, $\tan\theta=\frac{\text{Fe}}{\text{mg}}$

$=\frac{\text{kq}^2}{\text{r}}\times\frac{1}{\text{mg}}$

$\Rightarrow\frac{2}{\sqrt{1596}}=\frac{9\times10^9\times\text{q}^2}{(0.04)^2\times0.02\times9.8}$

$\Rightarrow\text{q}^2=\frac{(0.04)^2\times0.02\times9.8\times2}{9\times10^9\times\sqrt{1596}}$

$=\frac{6.27\times10^{-4}}{9\times10^9\times39.95}$

$=17\times10^{-16}\text{c}^2$

$\Rightarrow\text{q}=\sqrt{17\times10^{-16}}$

$=4.123\times10^{-8}\text{c}$

$\text{q}_1=\text{q}_2=2\times10^{-7}\text{c},\ \text{m}=100\text{g}$

$\text{l}=50\text{cm}=5\times10^{-2}\text{m},\ \text{d}=5\times10^{-2}\text{m}$

1. Now Electric force

$\text{F}=\text{K}\frac{\text{q}^2}{\text{r}^2}$

$=\frac{9\times10^9\times4\times10^{-14}}{25\times10^{-4}}\text{N}$

$=14.4\times10^{-2}\text{N}$

$=0.144\text{N}$

2. The components of Resultant force along it is zero, because mg balances $\text{T}\cos\theta$ and so also.

$\text{F}=\text{mg}=\text{T}\sin\theta$

3. Tension on the string

$\text{T}\sin\theta=\text{F},\ \text{T}\cos\theta=\text{mg}$

$\text{Tan}\theta=\frac{\text{F}}{\text{mg}}$

$=\frac{0.144}{100\times10^{-3}\times9.8}$

$=0.14693$

4. But $\text{T}\cos\theta=10^2\times10^{-3}\times10=1\text{N}$

$\Rightarrow\text{T}=\frac{1}{\cos\theta}=\sec\theta$

$\Rightarrow\text{T}=\frac{\text{F}}{\sin\theta},$

$\sin\theta=0.145369$

$\cos\theta=0.989378$

1. Let Q = charge on A & B Separated by distance d

q = charge on c displaced $\bot$ to -AB

So, force on $0=\overline{\text{F}}_\text{AB}+\overline{\text{F}}_\text{BO}$

But $\text{F}_\text{AO}\cos\theta=\text{F}_\text{BO}\cos\theta$

So, force on ‘0’ in due to vertical component.

$\overline{\text{F}}=\text{F}_\text{AO}\sin\theta+\text{F}_\text{BO}\sin\theta$ $|\text{F}_\text{AO}|=|\text{F}_\text{BO}|$

$=2\frac{\text{KQq}}{\Big(\frac{\text{d}}{2^2+\text{x}^2}\big)}\sin\theta$

$\text{F}=\frac{2\text{KQq}}{\Big(\frac{\text{d}}{2}\Big)^2+\text{x}^2}\sin\theta$

$=\frac{4\times2\times2\text{KQq}}{(\text{d}^2+4\text{x}^2)}\times\frac{\text{x}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{1}{2}}}$

$=\frac{2\text{kQq}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{3}{2}}}\text{x}$ = Electric force $\Rightarrow\text{F}\propto\text{x}$

2. When x << d

$\text{F}=\frac{2\text{kQq}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{3}{2}}}\text{x}$ x << d

$\Rightarrow\text{F}=\frac{2\text{kQq}}{\Big(\frac{\text{d}^2}{4}\Big)^{\frac{3}{2}}}\text{x}\Rightarrow\text{F}\propto\text{x}$

$\text{a}=\frac{\text{F}}{\text{m}}=\frac{1}{\text{m}}\Bigg[\frac{2\text{kQqx}}{\Big[\big(\frac{\text{d}^2}{4}\big)+\ell^2}\Bigg]$

So time period $\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$

$=2\pi\sqrt{\frac{\ell}{\text{a}}}$

$\text{E}=20\text{Kv/m}=20\times10^3\text{v/m},$

$\text{m}=80\times10^{-5}\text{kg},\ \text{c}=20\times10^{-5}\text{C}$

$\tan\theta=\Big(\frac{\text{qE}}{\text{mg}}\Big)^{-1}$ $\big[\text{T}\sin\theta=\text{mg},\ \text{T}\cos\theta=\text{qe}\big]$

$\tan\theta=\Big(\frac{2\times10^{-8}\times20\times10^3}{80\times10^{-6}\times10}\Big)$

$=\Big(\frac{1}{2}\Big)^{-1}$

$1+\tan^2\theta=\frac{1}{4}+1=\frac{5}{4}$ $\Big[\cos\theta=\frac{1}{\sqrt{5}},\sin\theta=\frac{2}{\sqrt{5}}\Big]$

$\text{T}\sin\theta=\text{mg}$

$\Rightarrow\text{T}\times\frac{2}{\sqrt{5}}=80\times10^{-6}\times10$

$\Rightarrow\text{T}=\frac{8\times10^{-4}\times\sqrt{5}}{2}=4\times\sqrt{5}\times10^{-4}$

$=8.9\times10^{-4}$

Force on the charge particle ‘q’ at ‘c’ is only the x component of 2 forces

So, $\text{F}_\text{on c}=\text{F}_\text{CB}\sin\theta+\text{F}_\text{AC}\sin\theta$

But $|\bar{\text{F}}_\text{CB}|=|\bar{\text{F}}_\text{AC}|$

$2\text{F}_\text{CB}\sin\theta$

$=2\frac{\text{KQq}}{\text{x}^2+\Big(\frac{\text{d}}{2}\Big)^2}\times\frac{\text{x}}{\Big[\frac{\text{x}^2+\text{d}^2}{4}\Big]^{\frac{1}{2}}}$

$=\frac{2\text{k}\theta\text{qx}}{\Big(\frac{\text{x}^2+\text{d}^2}{4}\Big)^{\frac{3}{2}}}$

$=\frac{16\text{kQq}}{\big(4\text{x}^2+\text{d}^2\big)^{\frac{3}{2}}}\text{x}$

For maximum force $\frac{\text{dF}}{\text{dx}}=0$

$\frac{\text{d}}{\text{dx}}\Bigg(\frac{16\text{kQqx}}{\big(4\text{x}^2+\text{d}^2\big)^{\frac{3}{2}}}\Bigg)=0$

$\Rightarrow\text{K}\begin{bmatrix}\frac{(4\text{x}^2+\text{d}^2)-\text{x}\bigg[\frac{3}{2\big[4\text{x}^2+\text{d}^2\big]^{\frac{1}{2}}}8\text{x}\bigg]}{\big[4\text{x}^2+\text{d}^2\big]^3}\end{bmatrix}=0$

$\Rightarrow\frac{\text{K}\big(4\text{x}^2+\text{d}^2\big)^{\frac{1}{2}}\Big[(4\text{x}^2+\text{d}^2)^3-12\text{x}^2\Big]}{\big(4\text{x}^2+\text{d}^2\big)^3}=0$

$\Rightarrow\big(4\text{x}^2+\text{d}^2\big)^3=12\text{x}^2$

$\Rightarrow16\text{x}^4+\text{d}^4+8\text{x}^2\text{d}^2=12\text{x}^2$ $[\text{d}^4+8\text{x}^2\text{d}^2=0]$

$\Rightarrow\text{d}^2=0$ $[\text{d}^2+8\text{x}^2=0]$

$\Rightarrow\text{d}^2=8\text{x}^2$

$\text{d}=\frac{\text{d}}{2\sqrt{2}}$