Question

Two pith balls of mass 10mg each are suspended by two threads from the same support are
charged identically. They move apart by 0.08m and threads make an angle 600 with each other.
Find the charge on each pith ball

Answer 1

Given,

Mass of the ball $= 10\text{ mg} = 1\times10^{-2}\text{ gm}$

Therefore, weight of the ball =$m\times{g}=1\times10^{-2}\times9.81\times10^{-4}=9.81\times10^{-6}\text{ N}$

In equilibrium, the net force is zero.

Therefor the summation of horizontal component of force is zero.

That means repulsive force is balanced by the sin component of the tension in the string.

$\sum{F_{H}}=0\\\\\Rightarrow{F}-Tsin\theta=0\\\\\Rightarrow{F}=Tsin\theta\\\\\Rightarrow{F}=Tsin30^{\circ}$................equation-1

The summation of vertical component of force is zero.

That means weight of the ball is balanced by the cos component of the tension in the string.

$\sum{F_{V}}=0\\\\\Rightarrow{F}-Tsin\theta=0\\\\\Rightarrow{mg}=Tcos\theta\\\\\Rightarrow{mg}=Tcos30^{\circ}\\\\\Rightarrow{T}=\frac{mg}{cos30^{\circ}}$ equation-2

Putting the value of T in equation-1 we get,

${F}=Tsin30^{\circ}\\\\\Rightarrow{F}=\frac{mg}{cos30^{\circ}}\times{sin}30^{\circ}\\\\\Rightarrow{F}=mg\times{tan}30^{\circ}\\\\\Rightarrow{F}=9.81\times10^{-6}\times\frac{1}{\sqrt{3}}$

The size of the repulsive electric forces acting on two pith ball with identical charge Q  is determined from Coulomb's law:

$F=\frac{1}{4\times\pi\times\xi_{0}}\times\frac{Q^{2}}{r^{2}}\\\\\Rightarrow{F}=k\times\frac{Q^{2}}{r^{2}}\text{ }[k\text{ is Constant and} k= 9\times10^{9}]\\\\\Rightarrow{F}={9\times10^{9}}\times\frac{Q^{2}}{(0.8)^{2}}\\\\\Rightarrow{Q}^{2}=F\times(0.8)^{2}\times\frac{1}{9\times10^{9}}\\\\\Rightarrow{Q}^{2}=9.81\times10^{-6}\times\frac{1}{\sqrt{3}}\times64\times10^{-4}\times\frac{1}{9\times10^{9}}$

$\\\\\Rightarrow{Q}^{2}=9.81\times10^{-6}\times\frac{1}{\sqrt{3}}\times64\times10^{-4}\times\frac{1}{9\times10^{9}}\\\\\Rightarrow{Q}^{2}=9.81\times0.577\times64\times\frac{1}{9}\times10^{-19}}\\\\\Rightarrow{Q}^{2}=40.27\times10^{-19}}\\\\\Rightarrow{Q}^{2}=4\times10^{-18}}\\\\\Rightarrow{Q}=2\times10^{-9}}$

The value of ${Q}=2\times10^{-9}}$ (approx)