A low - force capacitive actuator. A simple capacitive actuator can be made usin

Question

A low - force capacitive actuator. A simple capacitive actuator can be made using the configuration in Figure 5.101. Two coaxial tubes form a capacitor. The space between them contains a hollow tube made of Teflon that is free to move between the two cylinders as shown. Assume the moving part is at an arbitrary location x from the edge of the capacitor and the dimensions are as shown in the figure: Calculate the force the actuator can exert as a function of the applied voltage connected across the cylinders (positive to the outer cylinder, negative to the inner cylinder). Show from physical considerations that the motion can only be inward, regardless of polarity of the voltage applied. The capacitance of coaxial capacitors can be calculated directly but, as a first approximation, they may be treated as parallel plate capacitors using Equation (5.2), with S taken as the average between the areas of the outer and inner conductors, especially if the inner radius is not too small. The same applies to the electric field in the capacitor.

Explanation / Answer

1.Capacitance of coaxial cylindrical capacitor = C = L*2*pi*k*epsilon/ln(b/a)
For the given problem, the capacitors can be treated to be in parallel
co Ceff = [x + (c-x)*k]*2*pi*epsilon/ln(b/a)
Energy = 0.5CV^2 = 0.5*[x + (c-x)*k]*2*pi*epsilon/ln(b/a) * V^2 = [x + (c-x)*k]*pi*epsilon/ln(b/a) * v^2
noe, dE/dx = Force
Force = pi*epsilon(1 - k)*V^2/ln(b/a)

2. From the equation, we know, K > 1, so, Force is always negative ( i.e. it tries to reduce x), hence the force is always inward, and does not depend on sign of V as V^2 is always positive

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