Capacitance vs. Inverse Distance Graph's Slope

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The graph plotted shows capacitance on the y-axis and inverse distance on the x-axis, indicating a positive, increasing function. The equation derived from the linear fit is y=2.8e-13*x+6.1e-12, but there is uncertainty about whether a linear or quadratic fit is more appropriate. The slope is believed to relate to electric permittivity, yet doubts remain regarding its interpretation and significance. According to the theoretical model, the term ε*A corresponds to the slope value, while 1/d relates to the x-axis. Clarification on the numerical value of the slope and its implications is sought.
Ayda
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Homework Statement


I plotted a graph where Capacitance is at y-axis and Inverse Distance is at x-axis. It looks like a positive, increasing function.
7ba61bc988a2022d108dde02e1a1b54e-full.png

I am asked of a best fit, also an equation. But i am not sure whether to use quadratic or linear fit. I am also asked of what the slope corresponds to. The plate area of the capacitor was kept constant, we only changed distance.

Homework Equations


y=2.8e-13*x+6.1e-12 Is the equation of a linear slope.

The Attempt at a Solution


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I thought the slope would correspond to electric permittivity. But now i have big doubts about it. I also assumed the slope would be a linear fit, but I am not sure. Also, what does the numerical value of slope in the equation would mean? Any help would be very much appreciated. Thank you.
 
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The theoretical model is ##C = \epsilon * A/d##. (see https://www.electronics-tutorials.ws/capacitor/cap_4.html )
In the theoretical model, ##\epsilon * A## is related to your 2.8e-12, the ##1/d## is related to your x, and there is no constant term like your 6.1e-12.
 
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FactChecker said:
The theoretical model is ##C = \epsilon * A/d##. (see https://www.electronics-tutorials.ws/capacitor/cap_4.html )
In the theoretical model, ##\epsilon * A## is related to your 2.8e-12, the ##1/d## is related to your x, and there is no constant term like your 6.1e-12.
Thank you so much.
 
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