Chapter-25 - 1019 Chapter 25 1. (a) The capacitance of the...

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Unformatted text preview: 1019 Chapter 25 1. (a) The capacitance of the system is C q V = = = ∆ 70 20 35 pC V pF. . (b) The capacitance is independent of q ; it is still 3.5 pF. (c) The potential difference becomes ∆ V q C = = = 200 35 57 pC pF V. . 2. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV , and this is the same as the total charge that has passed through the battery. Thus, q = (25 × 10 –6 F)(120 V) = 3.0 × 10 –3 C. 3. For a given potential difference V , the charge on the surface of the plate is ( ) q Ne nAd e = = where d is the depth from which the electrons come in the plate, and n is the density of conduction electrons. The charge collected on the plate is related to the capacitance and the potential difference by q CV = (Eq. 25-1). Combining the two expressions leads to C d ne A V = . With 14 / / 5.0 10 m/V s s d V d V- = = × and 28 3 8.49 10 / m n = × (see, for example, Sample Problem 25-1), we obtain 28 3 19 4 2 (8.49 10 / m )(1.6 10 C)(5.0 10 14 m/V) 6.79 10 F/m C A-- = × × ×- = × . 4. We use C = A ε / d . (a) The distance between the plates is CHAPTER 25 1020 ( 29 ( 29 2 12 2 2 12 1.00m 8.85 10 C /N m 8.85 10 m. 1.00F A d C ε-- × ⋅ = = = × (b) Since d is much less than the size of an atom ( ∼ 10 –10 m), this capacitor cannot be constructed. 5. (a) The capacitance of a parallel-plate capacitor is given by C = ε A / d , where A is the area of each plate and d is the plate separation. Since the plates are circular, the plate area is A = π R 2 , where R is the radius of a plate. Thus, ( 29 ( 29 2 12 2 2 10 3 8.85 10 F m 8.2 10 m 1.44 10 F 144pF. 1.3 10 m R C d π ε π---- × × = = = × = × (b) The charge on the positive plate is given by q = CV , where V is the potential difference across the plates. Thus, q = (1.44 × 10 –10 F)(120 V) = 1.73 × 10 –8 C = 17.3 nC. 6. (a) We use Eq. 25-17: C ab b a =- = ×- = ⋅ 4 40 0 38 0 8 99 10 40 0 38 0 84 5 9 b ε . . . . . . mm mm mm mm pF. N m C 2 2 b gb g d ib g (b) Let the area required be A . Then C = ε A /( b – a ), or ( 29 ( 29 ( 29 ( 29 2 2 2 12 C N m 84.5pF 40.0mm 38.0mm 191cm . 8.85 10 C b a A ε- ⋅-- = = = × 7. Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4 π ε R to find the capacitance. When the drops combine, the volume is doubled. It is then V = 2(4 π /3) R 3 . The new radius R' is given by ( 29 3 3 4 4 2 3 3 R R ′ = ⇒ p p ′ = R R 2 1 3 . The new capacitance is 1 3 4 4 2 5.04 . C R R R ε ε ε ′ ′ = = = p p p With R = 2.00 mm, we obtain ( 29 ( 29 12 3 13 5.04 8.85 10 F m 2.00 10 m 2.80 10 F C π--- = × × = × ....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.

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Chapter-25 - 1019 Chapter 25 1. (a) The capacitance of the...

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