The discussion revolves around calculating the desired length of capacitor plates to achieve a specific capacitance of 0.95 pF, given the dimensions and materials involved in the capacitor's construction, including aluminum sheets and paper as a dielectric.
Participants are actively engaging with the problem, questioning the definitions of variables and the implications of the materials' properties. Some guidance has been provided regarding the equations and the importance of considering the number of layers of paper in the calculations.
There is some ambiguity regarding the thickness of the aluminum foil and how it affects the separation distance, as well as the interpretation of the thickness of the paper in relation to the overall setup of the capacitor.
negitron said:Seems like you might need an equation which relates plate area, separation and dielectric constant to the total capacitance, yes? And, of course, some super-basic geometry.
negitron said:Presumably for your coursework, you have covered these. There is only one which applies here. See if you can't figure it out and post it here; I'll tell you if you have the correct equation. Then, you can take the next step.
negitron said:In your equation above, A is the area, specifically, the area of overlap between the two plates (which in your problem is the area of either plate since it's implicitly assumed they overlap completely) and it is stated is square meters. C, of course, is the capacitance in Farads, E is the permitivitty of free space (8.854x10^-12 F/m ), k is your dielectric constant and d is the separation between the two plates (in meters). Now, apply your algebra to isolate A on one side of the equation.
Apprentice123 said:A is length of the leaves X width ? What is the value of d?
negitron said:Yes, exactly. And since the width is known, the rest is easy
What do you think the value of d is? What must determine that value in the problem setup?
negitron said:You're getting there. Since charge collection is a surface phenomenon only, the separation distance need only concern itself with the inner surfaces of the plates. It's how far apart these surfaces are that matters.
And remember to keep in mind how many layers of paper are being called for here.
negitron said:No, there is a reason the problem does not specify the thickness of the foil.
negitron said:Almost, but you're forgetting to take into account the number of layers of paper. Your answer is correct for a single layer of the stated paper.
negitron said:I'm pretty sure the problem as stated gives the thickness of one sheet of paper and specifies that two such sheets are used. Is that not the case? You're in a better position than I to definitively answer that, as I don't have (and probably couldn't read) the problem in its original language.
To be clear, if the problem gives the total thickness of the two sheets of paper, your answer is correct; if it gives only the thickness of one you obviously must multiply by 2.