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I’m trying to become an electrical engineer and I was given a problem in school (calculus-based physics) to work on. Basically, a parallel plate capacitor is charged to the same potential difference as the battery that is connected to it. Then, a metal slab was inserted between the two plates (after the battery is removed) and the new potential difference (V) across the capacitor was measured. Next, various metal slabs with different thicknesses (cm) replace the original, again measuring the new potential difference across the capacitor. I was instructed to graph potential difference (V) as a function of slab thickness and got a linear relationship with a slope of about -2.5 V/cm.

Now, I’m tasked with finding charge density (σ), which is charge/area, and the distance between the two plates. My professor insisted I must use the slope to do so. How do I do this?

To be quite honest, I’m quite lost. I’m not too familiar with the capacitance equations but I know that Charge = Capacitance * V and Capacitance = KEoArea/distance, even though I’m unsure if these relate. I derived V without the metal slab to be σd/2Eo but I don’t think I can use this with a metal slab now inserted. Any help is appreciated! Hopefully someone will be able to find the answer!

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Yes, the relationship of voltage and slab thickness is linear.

The metal slab effectively brings the plates closer together. You can consider the plates separated by d to be separated by d-a when you insert a slab of thickness a.

As long as the slab does not touch the plates, the charge remains constant.

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  • \$\begingroup\$ Thank you for that! How would I use the slope to find this charge density, though? \$\endgroup\$ – physicskid575 Mar 4 '20 at 5:45
  • \$\begingroup\$ Actually, you know the slope is in V/m. Charge density is Q/m^2 and V=Q/C and you know the parallel plate capacitor formula, so you can work this out. You might have to take a derivative. \$\endgroup\$ – Spehro Pefhany Mar 4 '20 at 5:49
  • \$\begingroup\$ The prompt says to “use the line of best fit (-2.5 V/cm) to find the charge density on the plates” and also to use the same line of best fit to “find the distance between the plates”. Is it even possible to just use the relationship between potential difference and the slab’s thickness to find this information? \$\endgroup\$ – physicskid575 Mar 4 '20 at 5:56
  • \$\begingroup\$ What does the intercept mean? Think about the physical interpretation of the slope and the intercept. \$\endgroup\$ – Spehro Pefhany Mar 4 '20 at 5:57
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    \$\begingroup\$ I wasn’t provided an intercept even though it should be the potential difference without the slab. All I have is that m=-2.5 V/cm. I think I’m starting to understand what you said. I’m working through it now. \$\endgroup\$ – physicskid575 Mar 4 '20 at 6:00

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