Capacitance varying with temperature.

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Homework Help Overview

The problem involves designing a capacitor that maintains constant capacitance despite temperature fluctuations. It focuses on a parallel-plate capacitor with plastic spacers and requires the derivation of the relationship between capacitance and temperature changes, as well as considerations regarding the thermal expansion of materials involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss differentiating the capacitance formula with respect to temperature and the implications of area and plate separation changes. There are questions about the application of differentiation rules and the effects of temperature on the capacitor's components.

Discussion Status

Some participants have provided guidance on using differentiation techniques to analyze the problem, while others have requested visual aids to better understand the setup. There is an ongoing exploration of the relationship between area change and plate separation in the context of thermal expansion.

Contextual Notes

Participants are considering the assumption that the plates are ultra-thin and discussing the impact of the spacers on the capacitance. There is also mention of needing to equate the rate of change of capacitance to zero for the design requirements.

Jonnyto
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Homework Statement


A capacitor is to be designed to operate, with constant capacitance, in an environment of fluctuating temperature. As shown in Fig. 23, the capacitor is a parallel-plate type with plastic "spacers" to keep the plates aligned. (a) Show that the rate of change of capacitance C with temperature T is given by

dC/dT= C[dA/AdT - dx/xdt]

where A is the plate area and x the plate separation. (b) If the plates are aluminum, what should be the coefficient of thermal expansion of the spacers in order that the capacitance not vary with temperature? (Ignore the effect of the spacers on the capacitance)

Homework Equations


C=Q/V C=εA/d


The Attempt at a Solution



I have no idea how to approach the problem. So far I can only tell that the temperature is causing the area change and I'm assuming that it's also causing the plates to move away or towards each other.
 
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It is known that [itex]C=ε_0\frac{A}{d}[/itex] right? Now differentiate this with respect to temperature (using chain rule to obtain the required result; If you have not learned chain rule yet, go through your maths textbook. I'm sure you'll get it there.)

Assuming that the plates are ultra-thin (i.e. they have negligible thickness) x doesn't change with temperature. So, the spacers should simply prevent A from changing.

P.S. Could you please upload the diagram ( a rough sketch is fine). I'm not able to visualize the spacers.
 
Okay I have attached the diagrams. How would I go about differentiating with respect to Capacitance and Temperature.
 

Attachments

  • spacer.JPG
    spacer.JPG
    7.8 KB · Views: 526
Jonnyto said:
Okay I have attached the diagrams. How would I go about differentiating with respect to Capacitance and Temperature.

Just so the I can answer to your queries better, have you learned about product rule, quotient rule and chain rule in differentiation?
 
Yes I have.
 
In that case you need to differentiate ε0A/x w.r.t. T. Use chain rule and quotient rule here.

For the capacitance to not change with temperature, the expression for dC/dT should be equated to zero (because capacitance should remain constant according to your question).

Thus you can obtain a relation between the infinitesimal change in area w.r.t. T (dA/dT) and the infinitesimal change in x with respect to temperature(dx/dt). You can find dA/Adt using tables (that is simply the coefficient of thermal expansion of area). Thus you can find dx/xdT.

Edit: dx/xdT is the coefficient of linear expansion of the plastic.
 
Sunil Simha said:
In that case you need to differentiate ε0A/x w.r.t. T. Use chain rule and quotient rule here.

For the capacitance to not change with temperature, the expression for dC/dT should be equated to zero (because capacitance should remain constant according to your question).

Thus you can obtain a relation between the infinitesimal change in area w.r.t. T (dA/dT) and the infinitesimal change in x with respect to temperature(dx/dt). You can find dA/Adt using tables (that is simply the coefficient of thermal expansion of area). Thus you can find dx/xdT.

Edit: dx/xdT is the coefficient of linear expansion of the plastic.
Ahh! Thank you so much! I had no idea about the linear expansion.
 

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