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Plane capacitor simulation

# Plane capacitor

This is an example of the plane capacitor simulation, performed with QuickField software.

**Problem Type:**

Plane problem of AC conduction.

**Geometry:**

**Given:**

Relative permittivity of substrate ε = 10.

Conductivity of substrate *g* = 10^{-10} S/m;

Voltage *U* = 5 V,

Frequency *f* = 1000 kHz.

Loss tangent *tan(δ)* = 0.01.

**Solution:**

Capacitor with non-ideal dielectric can be replaced by electric circuit with ideal capacitor *C* and resistivity *R* connected in parallel.

Current *I* has two components: active *I*_{A} and reactive *I*_{R}.

Active current due to dielectric conductivity *g* is very low. It could be calculated if we run the analysis with actual value of electric conductivity *g*.

Dielectric loss cause the electric current active component *I*_{A} to increase. It could be calculated if we run the analysis with effective value of electric conductivity, which is

*g*_{apparent} = 2π*f*·εε_{0}·tan(δ) = 2·3.142·1000000·10·(8.854·10^{-12})·0.01 = 5.56·10^{-6} S/m.

**Results:**

Electric current vectors and electric potential distribution in plane capacitor

Loss tangent tan(δ) = |*P*_{A} / *P*_{R}| = 0.0000695 / 0.00695 = 0.01.

View simulation report in PDF.

Download simulation files (files may be viewed using any QuickField Edition).