I start to learn electrostatic and I have some problem with finding the upper and lower boundaries of my parametric variable that I used to represent the graph of the potential surfaces of two positive charges place in the z axis, (0,0,a) and (0,0,-a). Just to say, i'm not a student, i study physics for my own pleasure, and my level in math is a little rusted. Sorry if the question is dumb!
I'm not a very good english speaker, so here the math:
So I have
$$ r_-=\sqrt{x^2+y²+(z-a)²} $$ $$ r_+=\sqrt{x^2+y²+(z+a)²} $$
My equation is $$ \phi(X)=\frac{Q}{4\pi\epsilon_0}(\frac{1}{r_-}+\frac{1}{r_+}) $$
I put $\rho=\sqrt{(\frac{x}{a})²+(\frac{y}{a})²}$, $\alpha=\frac{z}{a}$ and $V=\frac{4\pi\epsilon_0a}{2Q}\phi(X)$. With that i have :
$$ V=\frac{1}{2}(\frac{1}{\sqrt{\rho²+\alpha²+1-2\alpha}}+\frac{1}{\sqrt{\rho²+\alpha²+1+2\alpha}}) $$
To parametrise this equation I put $\left|u \right|<1$ and also :
$$ \frac{1}{\sqrt{\rho²+\alpha²+1-2\alpha}}= (1+u)V=\frac{a}{r_-} $$ $$ \frac{1}{\sqrt{\rho²+\alpha²+1+2\alpha}}= (1-u)V=\frac{a}{r_+} $$
Little explanation of why i'm doing that. From the expressions i just wrote, i have:
$$ V=\frac{1}{2}(\frac{a}{r_-} + \frac{a}{r_+}) \Leftrightarrow V=\frac{1}{2}((1+u)V + (1-u)V) $$
Like $\left|u \right|<1$, the equality is respected: $$ \lim_{u\rightarrow 1} V = \frac{1}{2}(2V+0)= V $$ $$ \lim_{u\rightarrow -1} V = \frac{1}{2}(0+2V)= V $$
From that, i'm able to get :
$$ \alpha=\frac{1}{V²(1-u²)²} $$ $$ \rho=\sqrt{(\frac{1}{V²(1-u²)²}-1)(1-\frac{u²}{V²(1-u²)²})} $$
And i have lot of difficulties here to find the upper and lower boundaries of $u$. I'm stuck here :
$$ \frac{1}{V(1-u²)}\geq 1 \geq \frac{\left|u \right|}{V(1-u²)} $$
But form that point I don't know how to find $u_{sup}$ and $u_{inf}$.
If anybody can help me, thanks!