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---?

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 `AB`
2. Tension `T` in the silk thread.
3. Weight acting downwards `mg`
   ![http://vidyadrishti.com]()

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^@-theta/2)`. (Sum of all three angles of a triangle `=180^@`.)

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

`AB=2xx[0.3xxsin(theta/2)]`
`AB=0.6sin(theta/2)m`

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

`F = k_e (|q_A q_B|)/ (AB)^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`
`=>F = (6.24xx10^-3)/ (sin^2(theta/2))`

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
`F/(sin(180-theta))=(mg)/(sin(90+theta/2))=T/(sin(90+theta/2))`

Using first equality and simplifying we get

`F/(sintheta)=(mg)/(cos(theta/2))`

Taking `g=9.81ms^-2`, inserting various values and rewriting `sin theta` in terms of half angle we get
`((6.24xx10^-3)/ (sin^2(theta/2)))/(2sin(theta/2)cos(theta/2))=(0.1xx9.81)/(cos(theta/2))`
`=>(6.24xx10^-3)/(2sin^3(theta/2))=0.1xx9.81`
`=>sin^3(theta/2)=(6.24xx10^-3)/(2xx0.1xx9.81)`
`=>sin^3(theta/2)=0.00318`
`=>sin(theta/2)=0.14706`
`=>(theta/2)=sin^-1 0.14706`
`=>theta=16.9^@`

It is clear from the figure that `Delta OAB` is an isosceles triangle, so

`2beta+theta=180^@`
`=>beta=90^@-theta/2`

Also,

`alpha+beta=90^@`
`=>alpha=90^@-(90^@-theta/2) = theta/2 ........................(1)`

From figure it is clear that in equilibrium position,

`mg*sin(theta)=F_e*cos(alpha)`

`=>mg*sin(2alpha)=F_e*cos(alpha)`

`=>2mg*sinalpha*cancelcolor(red)(cosalpha)=F_e*cancelcolor(red)(cos(alpha))`

`=>2mg*sinalpha=k_e(Q_AQ_B)/r^2=k_e(Q^2)/r^2...................(2)`

Now from `DeltaOAB`,

`costheta=(l^2+l^2-r^2)/(2*l^2)=1-(r^2)/(2l^2)`

`=>r^2=2l^2(1-costheta)=2l^2(1-cos(2alpha))`

Put the value of `r^2` in equation `(2)`,

`=>2mg*sinalpha=k_e(Q^2)/(2l^2(1-cos(2alpha)))=k_e(Q^2)/(2l^2(2-2cos^2alpha))`

`=>2mg*sinalpha=(k_eQ^2)/(4l^2sin^2alpha)`

`=>sinalpha=root(3)((k_eQ^2)/(8mgl^2))=root(3)((8.9875517873681764*10^9*(5*10^-7)^2)/(8*0.1*9.80665*(0.3)^2))= sin(theta/2)`

`=>theta=2*sin^-1(root(3)((8.9875517873681764*10^9*(5*10^-7)^2)/(8*0.1*9.80665*(0.3)^2)))`

`=>theta~~16.9164^@~~0.29525` rad

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 `AB`
2. Tension `T` in the silk thread.
3. Weight acting downwards `mg`
   ![http://vidyadrishti.com]()

As three forces are in equilibrium, sum of vertical components of `T and F` are equal and opposite to `mg` and Horizontal components of `T and F` are equal and opposite. From the figure we see that `F` makes an angle`=(90-theta/2)` with the vertical and `T` makes an angle `=theta` with the vertical.

We get
`Tcostheta+Fcos(90-theta/2)=mg`
`=>Tcostheta+Fsin(theta/2)=mg` ..... (1)

`Tsintheta=Fsin(90-theta/2)` .....(2)
From (2)
`=>T=1/sin theta Fcos(theta/2)`
Inserting this value of `T` in (1) we get

`(1/sin theta Fcos(theta/2))costheta+Fsin(theta/2)=mg`
`=>F (cos(theta/2)costheta+sin thetasin(theta/2))=mgsin theta`

Using the trigonometric identity we get

`F cos(theta-theta/2)=mgsin theta`
`=>Fcos(theta/2)=mgsin theta`

Proceed as in other solution