Experiment 5: The Simulation of Electrostatic Field
Name: Snober Batool
Student ID: 2111W020
Objectives
1) To understand the theoretical basis of simulation methods.
2) To strengthen the conceptual understanding of electric field strength and electric potential,
and to understand the relationship between electric field lines and equipotential lines.
3) To indirectly measure and delineate the electrostatic field of a long coaxial cylindrical cable
by using a simulation method.
Instruments"
Quick Field Student" simulation software is used in this experiment
Principle (summarize using less than 500 words)
1. Electrostatic Field
Electrostatic fields arise from a potential difference or voltage gradient. Electrostatic fields are
electric fields that do not change with time. Such fields are present when systems of charged matter
are stationary, or when electric currents are unchanging.
For an electrostatic field, the electric field strength satisfies the following integral relations in the
region without free charges:
∯s E.ds = 0
∮l E. dl = 0
2. Steady Current Field
Steady current fields arise when stationary charges produce magnetic fields that are constant in time;
hence the term magnetostatics is used. By steady currents we mean a flow of charge which has been
going forever, never increasing or decreasing.
For a steady current field, its current density also satisfies analogous integral relations in the region
without current sources.
∯s J.ds = 0
∮l J.dl = 0
3. The principle of simulation
Because the steady current field and the electrostatic field analogous to it are similar in terms of
spatial configuration, as long as the we can make the boundary conditions stable and the medium is
homogenous, the following connections can be established at any point in the region of interest:
Usteady = Uelectrostatic or Esteady = Electrostatic
Next, we show this equivalence mathematically using the electrostatic field of coaxial cylindrical
cable and the steady current field simulating it.
1
Electric field strength is:
Eq. (1)
The electric potential is:
Eq. (2)
The electric field strength of every point whose distance from the axis O is the radius r is:
Eq. (3)
The electric field strength E' of the steady current field is:
Eq. (4)
The distribution of the steady current field E' is equal to that of the electrostatic field E because
Eq. (5)
In reality, it is impossible to calculate the electrostatic field of an arbitrary charged body and the
potential distribution function of its simulating field. Only if the distribution of charges is
homogeneous and the shape of the charged body is symmetrical, can the electric field be determined
according to theoretic formula.
The simulation method is used under a certain condition and cannot be extended freely, otherwise
absurd conclusion may be arrived at. When using a steady current field to simulate an electrostatic
field, the following conditions should be satisfied:
1) The electrode geometry in the steady current field must be the same as the geometry of
the simulated charged body.
2) The conducting medium in the steady current field should be a poor conductor. In
particular, it must meet the following requirement:
 electrode
  conductor
Only when this requirement is satisfied can it be ensured that the surface of the electrode in the
steady current field is an approximate equipotential surface. Furthermore, the conductivity must be
distributed homogenously.
3) The boundary condition of the electrode system used for simulation should be the same as
that of the charged body to be simulated.
2
Procedure
1) Data recording
Table 1. The distance from the center to each point of the corresponding potential
Angle
2V
4V
6V
8V
0°
1.44
0.82
0.46
0.26
45°
1.43
0.82
0.45
0.25
90°
1.42
0.81
0.47
0.26
135°
1.41
0.82
0.47
0.25
180°
1.44
0.82
0.47
0.27
225°
1.43
0.81
0.45
0.25
270°
1.42
0.8
0.46
0.25
315°
1.43
0.81
0.45
0.25
average r
1.4275
0.814
0.46
0.255
Voltage
2.00
4.00
6.00
8.00
Lnr
0.36
-0.21
-0.78
-1.37
2) Data processing
Picture 1. Drawing the U-lnr graph.
U-lnr Graph
0.50
0.00
lnr
0.00
5.00
10.00
-0.50
-1.00
-1.50
y = -0.2869x + 0.9361
U(v)
3
Picture 2. Drawing the equipotential lines and the electric field lines.
3) Data calculation
ra = 0.15 cm, Ua = 10V,
rb = 2.50 cm, Ub = 0V,
= ln {(2.50/0.15)}÷(0-10) = ln(16.6)÷(-10) = -0.281
kmeasuring = -0.2869 ≈ 0.287
Relative error Ek =| ktheory－kmeasuing|/ ktheory ×100% = |{-0.281-(-0.287)}|÷(-0.281)×100%
= -2.135%
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