Introduction: The experiments for this lab are about how to find Dielectric Relative Permittivity and Dielectric Breakdown Strength of polypropylene. From the outcome of these two experiments, the Energy density of polypropylene could be found. The only way to find energy density for this lab is to use Dielectric materials and be able to created capacitors. Capacitors are an object that can store a certain amount of energy for a circuit. Capacitors are used in every commercial grade items such as pacemakers to military devices like a laser rail gun. The Military uses capacitors for pulse power storage of energy and capacitors are able to release it when needed for weapons and devices as radar and nuclear weapons. The Lab experiment stores
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In this case the capacitor is a parallel plate which means two metal conductors form dielectric materials that are stuck together using a glue type material. In our experiment the metal is a polypropylene film with the metal side and is connected using silicone. From the capacitor’s capacitance, relative permittivity could be found. In the second experiment, the Dielectric Breakdown Strength is found. This is done by using a high voltage source (60 kV) and is connected to a piece of polypropylene film on the metal side. This causes a breakdown and that can be seen on a graph to found the voltage when it had its …show more content…
The data above is my data found in the lab and later on my class data to solve the Weibull Plot for error. To find Relative Permittivity use equation 1 and solve for εr as shown below.
Relative Permittivity:
Capacitance (1) εo = 8.85 x 10-12 s4A2m-3kg-1 (Permittivity of free space)
A = .00123 m2 (Area of capacitor) t = 8 x 10-6 m (Thickness of capacitor)
Capacitance=2.58*10^(-9) F (Capacitance)
Relative Permittivity (εr) = 1.896
After solving for Relative Permittivity from the first Experiment, is one step closer to finding Energy Density, but now the Breakdown Electric Field which is Breakdown Voltage Divided by Dielectric Thickness has to be found. So now Experiment 2 data will be used an now. The Breakdown Voltage is 5447 V. The energy density of a capacitor could now be solved by using equation 2.
Energy Density:
Ue=(1/2)r0E2 (2)
εo = 8.85 x 10-12 s4A2m-3kg-1 εr = 1.896
E = 680875000 V/m (Breakdown Electric Field)
After solving the equation for Ue, Ue = 3889640 J / m3
The next step in the lab is to plot the Weibull Distribution and the data will use every other student’s data from the lab to find the error and how different the values
In this case the capacitor is a parallel plate which means two metal conductors form dielectric materials that are stuck together using a glue type material. In our experiment the metal is a polypropylene film with the metal side and is connected using silicone. From the capacitor’s capacitance, relative permittivity could be found. In the second experiment, the Dielectric Breakdown Strength is found. This is done by using a high voltage source (60 kV) and is connected to a piece of polypropylene film on the metal side. This causes a breakdown and that can be seen on a graph to found the voltage when it had its …show more content…
The data above is my data found in the lab and later on my class data to solve the Weibull Plot for error. To find Relative Permittivity use equation 1 and solve for εr as shown below.
Relative Permittivity:
Capacitance (1) εo = 8.85 x 10-12 s4A2m-3kg-1 (Permittivity of free space)
A = .00123 m2 (Area of capacitor) t = 8 x 10-6 m (Thickness of capacitor)
Capacitance=2.58*10^(-9) F (Capacitance)
Relative Permittivity (εr) = 1.896
After solving for Relative Permittivity from the first Experiment, is one step closer to finding Energy Density, but now the Breakdown Electric Field which is Breakdown Voltage Divided by Dielectric Thickness has to be found. So now Experiment 2 data will be used an now. The Breakdown Voltage is 5447 V. The energy density of a capacitor could now be solved by using equation 2.
Energy Density:
Ue=(1/2)r0E2 (2)
εo = 8.85 x 10-12 s4A2m-3kg-1 εr = 1.896
E = 680875000 V/m (Breakdown Electric Field)
After solving the equation for Ue, Ue = 3889640 J / m3
The next step in the lab is to plot the Weibull Distribution and the data will use every other student’s data from the lab to find the error and how different the values