Chapter24_Notes

# Chapter24_Notes - PHY 184 Physics for Scientists Engineers...

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PHY 184 Physics for Scientists & Engineers 2 Chapter 24: Capacitors

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Exam 1, Section 002 distribution 9/23/2011 Physics for Scientists & Engineers 2, Lecture 13 2 0 20 40 60 80 100 120 0 5 10 15 20 25 30 35 number of students exam score (%) Problem: 1 2 3 4 5 6 7 % correct: 79 31 53 81 64 77 5
September 23, 2011 Physics for Scientists & Engineers 2, Lecture 13 3 Capacitance Concept: A capacitor consists of two separated conductors, usually called plates , even if these conductors are not simple planes. We will define a simple geometry and generalize from there. We will start with a capacitor consisting of two parallel conducting plates, each with area A separated by a distance d . We assume that these plates are in vacuum (air is very close to a vacuum).

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Parallel Plate Capacitor We charge the capacitor by placing a charge +q on the top plate a charge -q on the bottom plate Because the plates are conductors, the charge will distribute itself evenly over the surface of the conducting plates. What is the relationship between the charge on the plates and the potential difference between the plates. q q (using a battery) (The potential difference is V(+) - V(-) = V) + + + + + + - - - - - -
Parallel Plate Capacitor Consider two parallel conducting plates of area A separated by a distance d The upper plate has +q and the lower plate has q. The electric field between the plates points from the positively charged plate to the negatively charged plate. We will assume ideal parallel plate capacitors in which the electric field is constant between the plates and zero elsewhere. Real-life capacitors have fringe field near the edges.

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Parallel Plate Capacitor: The Field We can calculate the electric field between the plates using Gauss’ Law We take a Gaussian surface shown by the red dashed line Flux = EA (through bottom surface of the top plate only!) Enclosed charge = q 0 q E A enclosed 0 q A d E
Parallel Plate Capacitor: The Potential Now we calculate the electric potential difference (also called the ―voltage‖ or the ―potential‖ ) across the plates of the capacitor in terms of the electric field. We define the electric potential across the capacitor to be V We carry out the integral in the direction of the blue arrow.       A d q Ed ds E s d E V V V V V f i f i i f 0 180 cos

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Parallel Plate Capacitor: Capacitance Note: q and V are proportional to one another. We call the constant of proportionality the capacitance , C. … so the capacitance of a parallel plate capacitor consisting of two plates of area A separated by a distance d is Note that the capacitance depends only on geometrical factors and not on the amount of charge or the voltage across the capacitor.
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Chapter24_Notes - PHY 184 Physics for Scientists Engineers...

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