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Lesson_3.2_Printable_PPT

Lesson_3.2_Printable_PPT - Capacitance Circuits Eo Consider...

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Capacitance Circuits. Consider a capacitor that is charged with one plate positive and the other negative separated by air in between. + + + + + + + - - - - - - - E o E’ E=E o -E’ E o _ _ _ _ _ + + + + _ The electric field between the capacitor plates is E 0 What happens when a dielectric medium 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.
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+ + + + + + + - - - - - - - E o E’ E=E o -E’ E o _ _ _ _ _ + + + + _ E’ E 0 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 that opposes E o . The net field E between the capacitor plates now is : E = E 0 – E E The net field E is related to κ as: 0 E κ = If σ b represents the bound charge density σ f the free charge density: 0 0 f E σ ε = 0 ' b E σ ε =
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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? E o = 2.5 × 10 5 V.m -1 , Q f = 10 nC = 10 -8 C E = 1.2 × 10 5 V.m -1 , E = E 0 – E = 2.5 × 10 5 – 1.2 × 10 5 (a) 0 E E κ = = 2.08 = 1.3 × 10 5 V.m -1 0 E E κ = 5 5 2.5 10 1.2 10 × = × E = E 0 – E
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(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? E o = 2.5 x 10 5 V.m -1 , Q f = 10 nC = 10 -8 C E = 1.2 x 10 5 V.m -1 , (b) 0 0 f E σ ε = 0 f Q A ε = 0 f Q A ε = 0 0 f Q A E ε = 8 5 12 10 2.5 10 8.85 10 - - = × × × = 0.0045 m 2
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(c) What is the total induced charge on either side of the dielectric? E o = 2.5 x 10 5 V.m -1 , Q f = 10 nC = 10 -8 C (c) 0 ' b E σ ε = 0 b Q A ε = Q E = 1.3 × 10 5 , 0 ' b E A ε = Q b = E A ε 0 = 1.3 × 10 5 × 0.0045 × 8.85 × 10 -12 = 5.75 × 10 -9 C = 5.75 nC
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Capacitors connected in series The figure 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. C 1 C 2 C 3 Q -Q Q -Q Q -Q V V 1 V 2 V 3 The battery provides a total potential difference of V volts across the three capacitors.
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