Lesson_3.2 - Lesson 3.2 Capacitor Circuits 1. The effect of...

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Lesson 3.2 Capacitor Circuits 1. The effect of a dielectric Fig. 1 below shows a dielectric inserted between the plates of a capacitor. + + + + + + + - - - - - - - - - - - + + + + dielectric E o E’ E=E o -E’ Fig. 1 Let the field between the capacitor plates be E o without the dielectric. When a dielectric is introduced between the plates, although there are no free charges in the dielectric, displacement of positive and negative molecular charges occur in the dielectric near the capacitor plates. These are called bound charges because they cannot move about. The bound charge is negative near the positive plate of the capacitor, and positive near the negative plate of the capacitor. These bound charges produce an electric field E’ which is directed opposite to E o . Therefore, the net field E between the capacitor plates now is E = E o – E’ …1 This net field is related to the dielectric constant κ of the dielectric as: 0 E E κ = …2 We represent the bound charge density of the dielectric by σ b and the free charge density on the capacitor plates by σ f . Then we have 0 0 0 and ' f b E E σ ε = = Example 1 : Two parallel plates have charges Q and –Q. When the space between the plates is devoid of matter, the electric field is 2.5 x 10 5 V.m -1 . When the space is filled with a certain dielectric, the field is reduced to 1.2 x 10 5 V.m -1 . (a) What is the dielectric constant of the dielectric? (b) If Q = 10 nC, what is the area of the plates? (c) What is the total induced charge on either side of the dielectric? Solution: We have: E o = 2.5 × 10 5 V.m -1 , E = 1.2 × 10 5 V.m -1 , Q f = 10 nC. E = E o – E’ or, E’ = E o – E = (2.5 – 1.2)10 5 = 1.3 × 10 5 V.m -1
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(a) 0 5 5 2.5 1 2. 0 1.2 10 2.5 1 08 .2 E E κ = × × = = = (b) 0 0 0 9 5 2 2 1 0 0 / 10 10 2.5 10 8.85 10 0.0045 f f f Q A E m Q A E σ ε - - = = × = = = × × × (c) 0 ' b E = σ b = E’ ε o = 1.3 × 10 5 × 8.85 × 10 -12 = 1.15 μ C.m -2 Q b = A σ b = 0.0045 × 1.15 × 10 -6 = 5.17 nC 2. Combination of capacitors. (i) Capacitors in series: C 1 C 2 C 3 Q -Q Q -Q Q -Q V Fig. 2 V 1 V 2 V 3 Fig. 2 above shows three capacitors of capacitance C 1 , C 2 and C 3 connected in series and the system connected to a battery which is a source of potential. The battery provides a total potential difference of V volts across the three capacitors. This total potential will be divided among the three capacitors so
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that the potential difference across C 1 will be V 1 , across C 2 will be V 2 and across C 3 will be V 3 in such a way that V = V 1 + V 2 + V 3 . …(i) If Q is the amount of charge on the positive plate of capacitor 1, by induction a charge –Q will be on its negative plate. Therefore, a charge Q must exist on the positive plate of C 2 and –Q on its negative plate. Similarly a charge of Q will be on the positive plate of C 3 and –Q on its negative plate. In other words, charge on each of the capacitors is the same, namely Q.
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This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.

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Lesson_3.2 - Lesson 3.2 Capacitor Circuits 1. The effect of...

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