## Electrostatic

Outputs of the solution are:

- Electric Fluxdensity, Electric Fieldstrength, Electric Potential, Capacity, Electrode Voltage, Electrode Charge.

## Theory

Electrostatic Analysis

The electrostatic solutions describe the distribution of electric fields due to static charges and/or levels of electric potential.

The basis equations:

(1) rot **e** = 0

(2) div **d** = ρ

(3) **d** = ε **e**

Boundary conditions:

(4) **n** x **e** | Γ_{0e} = 0

(5) **n** * **d** | Γ_{0d} = 0

Electric scalar potential formulation:

(6) div ε grad v = - ρ with

(7) **e** = -grad v

Electrostatic weak v-formulation:

(8) ( -ε grad v, grad v’ )_{Ω} = 0,

for all v’ element of Ω

## Example

Point and Cylinder

Given is a charge Qf that is applied to a cylindrical face. The face starts at the position (0,0,0) has an radius of R and a length of d. Further there is another charge Q being applied to a point, that is positioned on the axis of the cylinder and has a distance of a (a>d) from the origin. The hole space around has a constant permittivity of ε.

Goal is to analyze for the potential φ along the axis of the cylinder using the following parameters:

Q = 1e-9 As

Qf= 1e-7 As

a = 0,35 m

d = 0,1 m

R = 0,1 m

ε = 8,85419e-12 F/m

## Results

Potential φ

The comparison gives good agreement between theory and numerical solution.