Question

Three particles, each of mass $1gm$ and carrying a charge $q$, are suspended from a common point by insulated massless string, each $100cm$ long. If the particles are in equilibrium and are located at the corner of an equilateral triangle of side length $3cm$. Calculate the charge $q$ on each particle. (Take $g=10m/s^2$)

Answer 1

Let $T$ be the tension in string $F=\frac{k{q}^2}{r^2}(k=\frac{1}{4\pi {\epsilon }_0})$

$F_2=F_3=\frac{k(q{)}^2}{(0.03{)}^2}=\frac{k(q{)}^2}{(0.03{)}^2}$

$T\cos \theta =mg$

$T\sin \theta =\frac{\sqrt{3}k{q}^2}{(0.03{)}^2}$

$\tan \theta =\frac{\sqrt{3}k{q}^2}{\left(0.03{)}^2\right(10\left)\right(0.001)}$

$\{m=1gmm=0.001kgg=10m/s^2$

$\sin \theta =\frac{\sqrt{3}}{100}$ $\frac{\sqrt{3}}{100}=\frac{\sqrt{3}(9\times {10}^9)(q{)}^2}{\left(0.03{)}^2\right(0.01)}$

$\theta =0.992$ $\frac{\left(0.01{)}^2\right(0.03{)}^2}{9\times {10}^9}=q^2$

$\tan \theta =1.732\times {10}^{-2}$ $\frac{(3\times {10}^{-4}{)}^2}{9\times {10}^9}=q$

$\tan \theta \approx \frac{\sqrt{3}}{100}$ $q\approx \sqrt{\frac{{10}^{-16}}{10}}$

$q\approx 0.316\times {10}^{-8}C$