2015年AP电磁学简答题真题+答案+PDF下载
E&M.1.
A parallel-plate capacitor is constructed of two parallel metal plates, each with area A and separated by a distance D. The plates of the capacitor are each given a charge of magnitude Q, as shown in the figure above Ignore edge effects.
(a)
i. On the figure above, draw an arrow to indicate the direction of the electric field between the plates.
ii. On the figure above, draw an appropriate Gaussian surface that will be used to derive an expression for the magnitude of the electric field E between the plates.
iii. Using Gauss’s law and the Gaussian surface from part (a)-ii, derive an expression for the magnitude of the electric field E between the plates. Express your answer in terms of A, D, Q, and physical constants, as appropriate.
The space between the plates is now filled with a dielectric material that is engineered so that its dielectric constant varies with the distance from the bottom plate to the top plate, defined by the x-axis indicated in the diagram above. As a result, the electric field between the plates is given by 
where K0 is a positive constant. Express all algebraic answers to the remaining parts in terms of A, D, Q,K0, x, and physical constants, as appropriate.
(b) Determine an expression for the dielectric constant k as a function of x.
(c)i. Write, but do NOT solve, an equation that could be used to determine the potential difference V between the plates of the capacitor.
ii. Using the equation from part (c)-i, derive an expression for the potential difference VD-V0, where VD is the potential of the top plate and V0 is the potential of the bottom plate.
(d) Determine the capacitance of the capacitor.
(e) The energy stored in the capacitor that has a varying dielectric is UV. A second capacitor that has a constant dielectric of value K0 is also given a charge Q. The energy stored in the second capacitor is UC. How do the values of UV and UC compare?
——UV＜UC ——UV＞UC ——UV=UC
Justify your answer.
2015年AP电磁学简答题真题余下省略！
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