Electrostatic Clamps Using Dielectrics

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SUMMARY

Electrostatic clamps utilize dielectrics for holding workpieces during machining and microfabrication processes. The discussion focuses on a specific electrostatic clamp operating at 300 volts with a holding power of 2 atmospheres, using Mylar as the dielectric material. The challenge presented involves calculating the thickness of the Mylar insulator, given its dielectric constant of 3.2. Participants explore the application of parallel plate capacitor equations and the impact of a grounded plate on capacitance calculations.

PREREQUISITES
  • Understanding of electrostatics and dielectric materials
  • Familiarity with parallel plate capacitor equations
  • Knowledge of capacitance and electric field concepts
  • Basic skills in solving physics equations related to energy and force
NEXT STEPS
  • Research the effects of grounded plates on capacitance in electrostatic systems
  • Learn about the properties and applications of Mylar as a dielectric material
  • Explore advanced electrostatic clamp designs and their operational principles
  • Study the relationship between voltage, capacitance, and dielectric thickness in electrostatic applications
USEFUL FOR

Engineers and physicists involved in electrostatic applications, materials scientists researching dielectrics, and professionals in the fields of machining and microfabrication will benefit from this discussion.

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b]1. Homework Statement [/b]
Electrostatic clamps are used for holding workpieces while they are being machined, for holding silicon wafers during electron beam microfabrication, etc. They comprise an insulated conducting plate maintained at a potential of several thousand volts and covered with a thin insulating sheet. The workpiece or the wafer resets on the sheet and is grounded. It is advisabled to apply a film of oil to the sheet to prevent sparking.

One particular type operates at 300 volts and has holding power of 2 atmospheres. If the insulator is Mylar (Class-A; \epsilon_r=3.2, what is its thickness?


Homework Equations



F=-\frac{dW}{dx}

W=\frac{1}{2}CV^{2}

Q=CV


The Attempt at a Solution


Can I treat this system as a parallel plate conductor with a dielectric in the center? I don't see how to extract the thickness from this.

F==-\frac{dW}{dx}=\frac{1}{2}\frac{Q^2}{C^2}\frac{dC}{dx}=\frac{1}{2}V^2\frac{dC}{dx}

But the capacitance is constant C=\frac{Q}{\Delta V} meaning F becomes zero... which does not makes sense...

Is there another way to calculate the capacitance?
 
Last edited:
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I know the capacitance of a parallel plate configuration with dielectric filling the middle is C=\frac{A\epsilon_0\epsilon_r}{d}, but I also know that the grounded "plate" on top throws this off... I'm just not sure how
 

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