A particle A, having a charge of 5.0*10^-7C is fixed in a vertical wall. A second particle B of mass 100 g and having equal charge is suspended by a silk thread of length 30 cm from the wall. The point of suspension is 30 cm above the particle A---?

(continued) Find the angle of the thread with the vertical when it stays in equilibrium. Forces acting on the particle are (i) weight mg downward (ii) tension T along the thread (iii) electric force of repulsion F Coulomb's Law, Aug 9, 2017

Let $O$ be point of suspension of charged particle $B$. Charged particle $A$ is fixed and particle $B$ is held in equilibrium under forces as shown in the fig

1. Coulomb's Force of repulsion $F$ along $A B$
2. Tension $T$ in the silk thread.
3. Weight acting downwards $m g$ Let the thread make angle $\theta$ with the vertical.
We see that $\Delta A O B$ is an isosceles triangle.
Since its vertex angle $= \theta$, base angles are $= \left({90}^{\circ} - \frac{\theta}{2}\right)$. (Sum of all three angles of a triangle $= {180}^{\circ}$.)

If we drop a perpendicular from $O$ on $A B$, it divides the line $A B$ in two equal parts. Using trigonometry

$A B = 2 \times \left[0.3 \times \sin \left(\frac{\theta}{2}\right)\right]$
$A B = 0.6 \sin \left(\frac{\theta}{2}\right) m$

The magnitude of Coulomb's Force is found from the given expression

$F = {k}_{e} \frac{| {q}_{A} {q}_{B} |}{A B} ^ 2$
where ${k}_{e}$ is Coulomb's constant= 8.99×10^9 N m^2 C^-2

Inserting various values we get

F = 8.99×10^9xx (5.0xx10^-7xx5.0xx10^-7)/ (0.6sin(theta/2))^2
$\implies F = \frac{6.24 \times {10}^{-} 3}{{\sin}^{2} \left(\frac{\theta}{2}\right)}$

Using Lami's theorem which relates the magnitudes of three coplanar, concurrent and non-collinear forces, $T , F \mathmr{and} W$ keeping the charged particle $B$ in static equilibrium and angles between these we get
$\frac{F}{\sin \left(180 - \theta\right)} = \frac{m g}{\sin \left(90 + \frac{\theta}{2}\right)} = \frac{T}{\sin \left(90 + \frac{\theta}{2}\right)}$

Using first equality and simplifying we get

$\frac{F}{\sin \theta} = \frac{m g}{\cos \left(\frac{\theta}{2}\right)}$

Taking $g = 9.81 m {s}^{-} 2$, inserting various values and rewriting $\sin \theta$ in terms of half angle we get
$\frac{\frac{6.24 \times {10}^{-} 3}{{\sin}^{2} \left(\frac{\theta}{2}\right)}}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)} = \frac{0.1 \times 9.81}{\cos \left(\frac{\theta}{2}\right)}$
$\implies \frac{6.24 \times {10}^{-} 3}{2 {\sin}^{3} \left(\frac{\theta}{2}\right)} = 0.1 \times 9.81$
$\implies {\sin}^{3} \left(\frac{\theta}{2}\right) = \frac{6.24 \times {10}^{-} 3}{2 \times 0.1 \times 9.81}$
$\implies {\sin}^{3} \left(\frac{\theta}{2}\right) = 0.00318$
$\implies \sin \left(\frac{\theta}{2}\right) = 0.14706$
$\implies \left(\frac{\theta}{2}\right) = {\sin}^{-} 1 0.14706$
$\implies \theta = {16.9}^{\circ}$

Aug 9, 2017

$\theta \approx {16.9164}^{\circ} \approx 0.29525$ rad

Explanation: It is clear from the figure that $\Delta O A B$ is an isosceles triangle, so

$2 \beta + \theta = {180}^{\circ}$
$\implies \beta = {90}^{\circ} - \frac{\theta}{2}$

Also,

$\alpha + \beta = {90}^{\circ}$
$\implies \alpha = {90}^{\circ} - \left({90}^{\circ} - \frac{\theta}{2}\right) = \frac{\theta}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$

From figure it is clear that in equilibrium position,

$m g \cdot \sin \left(\theta\right) = {F}_{e} \cdot \cos \left(\alpha\right)$

$\implies m g \cdot \sin \left(2 \alpha\right) = {F}_{e} \cdot \cos \left(\alpha\right)$

$\implies 2 m g \cdot \sin \alpha \cdot \cancel{\textcolor{red}{\cos \alpha}} = {F}_{e} \cdot \cancel{\textcolor{red}{\cos \left(\alpha\right)}}$

$\implies 2 m g \cdot \sin \alpha = {k}_{e} \frac{{Q}_{A} {Q}_{B}}{r} ^ 2 = {k}_{e} \frac{{Q}^{2}}{r} ^ 2. \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$

Now from $\Delta O A B$,

$\cos \theta = \frac{{l}^{2} + {l}^{2} - {r}^{2}}{2 \cdot {l}^{2}} = 1 - \frac{{r}^{2}}{2 {l}^{2}}$

$\implies {r}^{2} = 2 {l}^{2} \left(1 - \cos \theta\right) = 2 {l}^{2} \left(1 - \cos \left(2 \alpha\right)\right)$

Put the value of ${r}^{2}$ in equation $\left(2\right)$,

$\implies 2 m g \cdot \sin \alpha = {k}_{e} \frac{{Q}^{2}}{2 {l}^{2} \left(1 - \cos \left(2 \alpha\right)\right)} = {k}_{e} \frac{{Q}^{2}}{2 {l}^{2} \left(2 - 2 {\cos}^{2} \alpha\right)}$

$\implies 2 m g \cdot \sin \alpha = \frac{{k}_{e} {Q}^{2}}{4 {l}^{2} {\sin}^{2} \alpha}$

$\implies \sin \alpha = \sqrt{\frac{{k}_{e} {Q}^{2}}{8 m g {l}^{2}}} = \sqrt{\frac{8.9875517873681764 \cdot {10}^{9} \cdot {\left(5 \cdot {10}^{-} 7\right)}^{2}}{8 \cdot 0.1 \cdot 9.80665 \cdot {\left(0.3\right)}^{2}}} = \sin \left(\frac{\theta}{2}\right)$

$\implies \theta = 2 \cdot {\sin}^{-} 1 \left(\sqrt{\frac{8.9875517873681764 \cdot {10}^{9} \cdot {\left(5 \cdot {10}^{-} 7\right)}^{2}}{8 \cdot 0.1 \cdot 9.80665 \cdot {\left(0.3\right)}^{2}}}\right)$

$\implies \theta \approx {16.9164}^{\circ} \approx 0.29525$ rad

Aug 9, 2017

Alternate solution steps if you don't want to use Lami's Theorem

Explanation:

Let $O$ be point of suspension of charged particle $B$. Charged particle $A$ is fixed and particle $B$ is held in equilibrium under forces as shown in the fig

1. Coulomb's Force of repulsion $F$ along $A B$
2. Tension $T$ in the silk thread.
3. Weight acting downwards $m g$ As three forces are in equilibrium, sum of vertical components of $T \mathmr{and} F$ are equal and opposite to $m g$ and Horizontal components of $T \mathmr{and} F$ are equal and opposite. From the figure we see that $F$ makes an angle$= \left(90 - \frac{\theta}{2}\right)$ with the vertical and $T$ makes an angle $= \theta$ with the vertical.

We get
$T \cos \theta + F \cos \left(90 - \frac{\theta}{2}\right) = m g$
$\implies T \cos \theta + F \sin \left(\frac{\theta}{2}\right) = m g$ ..... (1)

$T \sin \theta = F \sin \left(90 - \frac{\theta}{2}\right)$ .....(2)
From (2)
$\implies T = \frac{1}{\sin} \theta F \cos \left(\frac{\theta}{2}\right)$
Inserting this value of $T$ in (1) we get

$\left(\frac{1}{\sin} \theta F \cos \left(\frac{\theta}{2}\right)\right) \cos \theta + F \sin \left(\frac{\theta}{2}\right) = m g$
$\implies F \left(\cos \left(\frac{\theta}{2}\right) \cos \theta + \sin \theta \sin \left(\frac{\theta}{2}\right)\right) = m g \sin \theta$

Using the trigonometric identity we get

$F \cos \left(\theta - \frac{\theta}{2}\right) = m g \sin \theta$
$\implies F \cos \left(\frac{\theta}{2}\right) = m g \sin \theta$

Proceed as in other solution