# Question #98f20

Nov 25, 2015

$q = 6.66 \times {10}^{- 9} \text{C}$

#### Explanation: The weight acting downwards is balanced by the vertical component of the tension.

The electrostatic repulsion between the 2 charges is balanced by the horizontal component of the tension.

From the geometry of the triangle we can write:

$\sin 15 = \frac{r}{0.45}$

$\therefore r = 0.45 \times 0.259 = 0.11633$

$\therefore 2 r = 0.233 \text{m}$

Taking the vertical components:

$m g = T \cos 15$

$\therefore T = \frac{m g}{\cos 15} = \frac{2.8 \times {10}^{- 6} \times 9.8}{0.966}$

$\therefore T = 28.405 \text{N}$

Now for the horizontal components:

The force between the 2 charges is given by Coulomb's Law:

$F = \frac{1}{4 \pi {\epsilon}_{0}} . \frac{{q}^{2}}{r} ^ 2$

$\frac{1}{4 \pi {\epsilon}_{0}}$ is a constant and has the value $k = 9 \times {10}^{9} \text{m/F}$

$\therefore {F}_{E} = \frac{k {q}^{2}}{{0.233}^{2}} = T \cos 75$

Substituting in the value for $T$:

$\therefore \frac{k {q}^{2}}{0.233} ^ 2 = 28.405 \times {10}^{- 6} \times 0.259$

$\therefore {q}^{2} = \frac{28.405 \times {10}^{- 6} \times 0.259 \times {0.233}^{2}}{9 \times {10}^{9}}$

${q}^{2} = \frac{0.4 \times {10}^{- 6}}{9 \times {10}^{9}}$

${q}^{2} = 0.444 \times {10}^{- 16}$

$q = 6.66 \times {10}^{- 9} \text{C}$