Question

Three particles each of mass 1g and carrying a charge q are suspended from a common point by insulated massless strings each of 100cm long. if the particles are in equilibrium and are located at the corners of an equilateral triangle of sides of length 3 cm,calculate the charge q on each particle.

Answer 1

I think answer is 33.5

Answer 2

The charge of the given particle is $\bold{q=3.13 \times 10^{-9} C}$

Solution:

The three particles system shall be in equilibrium, if net force on the particles is zero.

Three forces are acting here weight, tension in the strings and electrostatic force.

Let particle be placed at PAB points

$F^{2}=F_{p a}^{2}+F_{p b}^{2}+2 F_{p a} \cdot F_{p b} \cdot \cos 60$

$F_{p a}=F_{p b}$

$=\frac{9 \times 10^{9} \times q^{2}}{0.03^{2}}$

$=10^{13} q^{2}$

$\mathrm{F}=\sqrt{3}\left(10^{13} q^{2}\right)$

$T \sin \theta=F$

$T \cos \theta=m g$

$\tan \theta=\frac{F}{m g}$

$=\frac{\left(\sqrt{3}\left(10^{13} q^{2}\right)\right.}{m g}$

Centroid of triangle = $\frac{a}{\sqrt{3}}$

Hence,

$\frac{a}{\sqrt{3}}=\frac{\left(\sqrt{3}\left(10^{13} q^{2}\right)\right.}{m g}$

$q=3.13 \times 10^{-9} C$