Question

A particle A having a charge of $5.0\times {10}^{-}7C$ 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 above the particle A. Find the angle of thread with the vertical when it stays in equilibrium.

Answer 1

Let the thread make angle $\theta$ with the vertical.

We see that $\Delta AOB$ is an isosceles triangle.

Since its vertex angle $=\theta ,$ base angles are $=({90}^{\circ }-\frac{\theta }{2}).$ . Sum of all three angles of a triangle $={180}^{\circ }$)

If we drop a perpendicular from $O$ on $AB,$ it divides the line $AB$ in two equal parts.

Using trigonometry

$AB=0.6\sin \left(\frac{\theta }{2}\right)m$

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

$F=k_e\frac{\left|q_A{q}_B\right|}{(AB{)}^2}$

where $k_e$ is Coulomb's constant $=8.99\times {10}^{9}N{m}^2{C}^{-2}$

Inserting various values we get

$F=8.99\times {10}^9\times \frac{5.0\times {10}^{-7}\times 5.0\times {10}^{-7}}{{\left(0.6\sin \left(\frac{\theta }{2}\right)\right)}^2}$

$\Rightarrow F=\frac{6.24\times {10}^{-3}}{{\sin }^2\left(\frac{\theta }{2}\right)}$

Using Lami's theorem which relates the magnitudes of three coplanar, concurrent

and non-collinear forces, $T,F$ and $W$ keeping the charged particle $B$ in static

equilibrium and angles between these we get

$\frac{F}{\sin (180-\theta )}=\frac{mg}{\sin (90+\frac{\theta }{2})}=\frac{T}{\sin (90+\frac{\theta }{2})}$

Using first equality and simplifying we get

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

Taking $g=9.81m{s}^{-2},$ inserting various values and rewriting sin $\theta$ in terms of half angle we get

$\frac{\frac{6.24\times {10}^{-3}}{{\sin }^2\left(\frac{\theta }{2}\right)}}{2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)}=\frac{0.1\times 9.81}{\cos \left(\frac{\theta }{2}\right)}$

$\Rightarrow \frac{6.24\times {10}^{-3}}{2{\sin }^3\left(\frac{\theta }{2}\right)}=0.1\times 9.81$

$\Rightarrow {\sin }^3\left(\frac{\theta }{2}\right)=\frac{6.24\times {10}^{-3}}{2\times 0.1\times 9.81}$

$\Rightarrow {\sin }^3\left(\frac{\theta }{2}\right)=0.00318$

$\Rightarrow \left(\frac{\theta }{2}\right)={\sin }^{-1}0.14706$

$\Rightarrow \theta ={16.9}^{\circ }$.