Question

An air-filled parallel-plate capacitor has plates with area A seperated by length d resulting in a capacitance value C. The plates were connected to a battery, and the capacitor was charged until the potential difference between the plates reached V, and finally the battery was removed.

CON CICLIC I DWT the pa p using only in ICT IV in ne prooi 177 Part MT NONSTON TO statement. E-Vod Correct! 17% Part (C) Enter an expression for the energy stored in the capacitor using only the parameters given in the problem statement. U - 1/2 CV? Correct! 17% Part (d) Eliminate Co and Vo from the expression in the previous step using the results of the earlier steps. Input the resulting expression for U. U - 1/2 (Ad) (d) Correct! V- A 17% Part (e) Enter an expression for the volume of the capacitor in terms of the parameters given in the problem statement Grade Summary Deductions 100% Potential 7 8 9 56 2 3 Submissions Attempts remaining: 10 ( per attempt) 1 VOC AR Submit Hint Feedback I give up! Hints: 29 deduction per hint. Hints remaining: Feedback: 2 deduction per feedback. 17% Part (1) The energy stored in the capacitor is represented by an upper-case U while the energy density is represented by a lower-case. Divide the expression from part (d) by the volume of the parallel-plate capacitor to obtain the energy density, and enter a simplified expression for u. (The result obtained generally applies to the energy density stored in an electric field even though we just derived it specifically in the context of a parallel-plate capacitor)

Need help with parts e and f specifically.

Answer 1

e)

volume of capacitor

Vol = area * separation

Vol = A d

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f)

u = (1/2) (eo A /d) (Ed)^2 / vol

u = (1/2) (eo A E^2 d) / (Ad)

u = (1/2) $\epsilon o$ * E^2

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