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Given $\tan^{-1}y = \tan^{-1}x + \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$ with $|x|<\dfrac{1}{\sqrt{3}}$, find $y$.
Use $\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)=2\tan^{-1}x$ (valid for $|x|<1$).
So $\tan^{-1}y = \tan^{-1}x + 2\tan^{-1}x = 3\tan^{-1}x$.
Using the triple angle formula: $\tan(3\theta)=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$:
$y = \tan(3\tan^{-1}x) = \dfrac{3x-x^3}{1-3x^2}$
According to Coulomb's Law, the force $F$ between two charges $q_1$ and $q_2$ separated by distance $r$ is:
$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$
Rearranging for permittivity:
$\epsilon_0 = \frac{q^2}{4\pi F r^2}$
Dimensional analysis:
- Charge $[q] = [AT]$
- Force $[F] = [MLT^{-2}]$
- Distance $[r] = [L]$
$[\epsilon_0] = \frac{[A^2 T^2]}{[M L T^{-2}] [L^2]} = [M^{-1} L^{-3} T^4 A^2]$
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