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CourseNotes.19 - Figure 13 A Parallel-Plate Capacitor...

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Unformatted text preview: Figure 13: A Parallel-Plate Capacitor conductor is zero, the only flux will be through the part of our Gaussian surface in between the two plates, and 2) Since the field is uniform, E · dA = E dA. Hence, E · dA = q ⇒ 0 E= q 0A where A is the area of one of the plates. Now that we know the field between the plates, we can calculate the potential between them. d d E · ds = V= 0 0 qd q ds = 0A 0A Now that we know the potential difference between the two plates, calculating the capacitance is simple. q q 0A C= = qd = V d A 0 Notice that, as stated before, the capacitance is simply a geometrical quantity. In the case of a parallel-plat capacitor, it is proportional to the area of each plate and inversely proportional to the distance between the two plates. 6.2.2 Cylindrical Capacitor Figure 14: A Cylindrical Capacitor We begin by assuming that there is a hypothetical charge q placed on the inner and outer conductor. Because of symmetry, the charge on the inner conductor will distribute itself uniformly on the outside edge of the cylinder. If we draw our Gaussian surface in the usual way and make the usual symmetry arguments, then the electric field between the plates is q E= r ˆ 2πrL 0 Hence, the potential difference between the plates is b V= a q dr = 2πrL 0 19 q ln 2πL 0 b a ...
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