I know that the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon S}{d}$. I am supposed to perform a linear regression to obtain $\varepsilon$, however it turns to be a quadratic relation, enter image description here.

My professor told us to justify it, so it might not be a mistake. It cannot be edge effect as the point that behave badly if we consider it a linear relation are those with smaller separation between plates. It cannot be caused by the saturation of the dielectric we used, as the electric field wasn't that strong. The problem isn't the absence of dielectric in a part of the separation between plates either.

We are using a parallel circular plate condensator. The surface of the plates is $S = \pi \left(\frac{0.255}{2}\right)^{2}$ m$^{2}$.

  • $\begingroup$ For a given plate separation you are getting a capacitance which seems to be too low. How did you keep the plates "pressed" onto the dielectric because with a dielectric of thickness $\frac 12 \,\rm mm$ a small air gap across parts of the plates will have a large effect. Also how flat were the plates? $\endgroup$ – Farcher Oct 30 '17 at 22:24
  • $\begingroup$ The capacitance is not that low, the diameter of the plates was $D = 0.255$ m, so without dielectric, at a distance of $0.5$ mm between the plates the capacitance would be around $900$ pF. And the plates were totally flat. The dielectric was well pressed, as it was held just by the capacitor's plates. $\endgroup$ – Alex Oct 30 '17 at 22:30
  • $\begingroup$ Well pressed by the weight of the top plate might not be good enough? If you can repeat the experiment distribute some weights on the top plate to see if that has any effect. $\endgroup$ – Farcher Oct 30 '17 at 22:36
  • $\begingroup$ It is not an experimental issue, as similar results have been obtained by other groups and we are supposed to find the reason why the quadratic regression works better and the linear term is a better stimation of $\varepsilon S$ than the slope of the linear regression. It does have a physical meaning, said my proffesor. $\endgroup$ – Alex Oct 30 '17 at 22:44
  • $\begingroup$ How was the capacitance measured? $\endgroup$ – Farcher Oct 30 '17 at 22:55

Since the only variation is the width of the dielectric material, and other effects have been already excluded, it could be a material that has a finite resistance. So there is a parallel resistor involved with a linear decreasing value the smaller the gap becomes reducing the apparent capacitance. This could shift the resonance frequency yielding an (apparently) non-linear dependency.

  • $\begingroup$ It would be nice if Alex disclosed the reason - the professor surely knows the answer. $\endgroup$ – xeeka Feb 19 '18 at 8:15
  • $\begingroup$ Alex, could you please publish the correct answer? Thanks. $\endgroup$ – xeeka Jan 30 '19 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.