Thats a great help thank you.
So I rerrange and integrate to give:
\frac{dV}{dρ}=\frac{C}{ρ^{2}}
\int\frac{C}{ρ^{2}}dρ=\frac{-C}{ρ}+D
So:
V(ρ)=\frac{-C}{ρ}+D
I know that at the optic axis (ρ0), V(ρ)=V0 which leaves me with:
V_{0}=\frac{-C}{ρ_{0}}+D
I'm struggling to work out what the...
Homework Statement
For some work I am doing I wish to be able to define the potential distribution as a function of the radius (ρ) between two concentric electrodes.
Homework Equations
One solution (from reliable literature) defines the varying radial potential as:
V(ρ)=2V0(ρ0/ρ...