jmtome2
Nov3-09, 01:09 PM
b]1. The problem statement, all variables and given/known data[/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?
2. Relevant equations
F=-\frac{dW}{dx}
W=\frac{1}{2}CV^{2}
Q=CV
3. 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?
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?
2. Relevant equations
F=-\frac{dW}{dx}
W=\frac{1}{2}CV^{2}
Q=CV
3. 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?