Question #98f20

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Answer 1 · 2015-11-25T14:10:17

$q = 6.66 \times 10^{- 9} C$ $q=6.66xx10^(-9)"C"$

MFDocs

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:

$sin 15 = \frac{r}{0.45}$ $sin15=r/0.45$

$:.r=0.45xx0.259=0.11633$

$:.2r=0.233"m"$

Taking the vertical components:

$mg=Tcos15$

$:.T=(mg)/(cos15)=(2.8xx10^(-6)xx9.8)/(0.966)$

$:.T=28.405"N"$

Now for the horizontal components:

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

$F=(1)/(4piepsilon_0).(q^2)/r^2$

$(1)/(4piepsilon_0)$ is a constant and has the value $k=9xx10^(9)"m/F"$

$:.F_E=(kq^2)/(0.233^2)=Tcos75$

Substituting in the value for $T$ :

$:.(kq^2)/0.233^2=28.405xx10^(-6)xx0.259$

$:.q^2=(28.405xx10^(-6)xx0.259xx0.233^2)/(9xx10^9)$

$q^2=(0.4xx10^(-6))/(9xx10^9)$

$q^2=0.444xx10^(-16)$

$q=6.66xx10^(-9)"C"$