Question 6.1: Current-carrying components in high-voltage power equipment ...

Since \rho_{v}\neq 0, we apply Poisson’s equation

\nabla^2 V=-\frac{\rho_{v}}{\varepsilon}

The boundary conditions V(z=0)=V_{o} and V(z=d)=0 show that V depends only on z (there is no \rho or \phi dependence). Hence

\frac{d^{2}V}{dz^{2}}=\frac{-\rho_{o}}{\varepsilon}

Integrating once gives

\frac{dV}{dz}=\frac{-\rho_{o}z}{\varepsilon} +A

Integrating again yields

V=-\frac{\rho_{o}z^{2}}{2\varepsilon}+Az+B

where A and B are integration constants to be determined by applying the boundary conditions. When z=0,  V=V_{o}

V_{o}=-0+0+B\rightarrow B=V_{o}

When z=d,  V=0

0=-\frac{\rho_{o}d^{2}}{2\varepsilon}+Ad+V_{o}

or

A=\frac{\rho_{o}d}{2\varepsilon}-\frac{V_{o}}{d}

The electric field is given by

E=-\nabla V=-\frac{dV}{dz}a_{z}=\left(\frac{\rho_{o}z}{\varepsilon}-A\right)a_{z}=\left[\frac{V_{o}}{d}+\frac{\rho_{o}}{\varepsilon}\left(z-\frac{d}{2}\right)\right]a_{z}

The net force is

F=\int_{v}\rho_{v}Edv=\rho_{o}\int dS\int_{z=0}^{d}Edz=\rho_{o}S\left[\frac{V_{o}z}{d}+\frac{\rho_{o}}{2\varepsilon}(z^{2}-dz) \right]|_{0}^{d}a_{z}

F=\rho_{o}SV_{o}a_{z}

The force per unit area or pressure is

\rho=\frac{F}{S}=\rho_{o}V_{o}=25\times 10^{-3}\times 22\times 10^{3}=550N/m^{2}