Question

Two light, conducting spheres, each 6 mm diameter and having a mass of 10 mg, are

suspended from the same point by fine insulated fibers of 50cm long. Due to electrostatic

repulsion the spheres are in equilibrium when 3 cm apart.

Answer 1

$sin\Big(\dfrac{\theta}{2}\Big)=\dfrac{1.5}{50}$

$\dfrac{\theta}{2}=sin^{-1}\Big(\dfrac{1.5}{50}\Big)$

$\theta=3.43\degree$

Balancing forces in x direction,

$Tcos(90\degree-1.715\degree)=F$

$T=\dfrac{F}{cos88.28\degree}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$

Balancing forces in y direction

$Tsin88.28\degree=mg$

$T=\dfrac{mg}{sin88.28\degree}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(2)$

From equation (1) and (2)

$\dfrac{F}{mg}=cot88.28\degree$

$F=mg\times cot88.28\degree$

$F=2.94\times10^{-3}\space N$

Let charges on both the sphere be $q$

Electrostatic force, $F=\dfrac{kq^2}{r^2}$

$q=\sqrt{\dfrac{(3\times10^{-2})^2(2.94\times10^{-3})}{9\times10^{9}}}$

$q=1.714\times10^{-8}\space C$