solutions_chapter18

# U b 5 1 2 u a 5 0099 j r b 5 0500 m k a 5 0 u a 5 1

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U b 5 1 2 U a 5 0.099 J. r b 5 0.500 m K a 5 0. U a 5 1 0.198 J, U 5 1 8.99 3 10 9 N # m 2 / C 2 2 1 1.20 3 10 2 6 C 21 4.60 3 10 2 6 C 2 0.250 m 5 0.198 J K 5 1 2 m v 2 . U a 1 K a 5 U b 1 K b . U 5 k qQ r . W B S A 5 q 1 V B 2 V A 2 5 1 2.50 3 10 2 9 C 21 2 704 V 2 1 2 737 V 22 5 8.2 3 10 2 8 J V B 5 kq 1 r 1 B 1 kq 2 r 2 B 5 1 8.99 3 10 9 N # m 2 / C 2 2 1 2.40 3 10 2 9 C 0.080 m 1 2 6.50 3 10 2 9 C 0.060 m 2 5 2 704 V V A 5 kq 1 r 1 A 1 kq 2 r 2 A 5 8.99 3 10 9 N # m 2 / C 2 0.050 m 1 2.40 3 10 2 9 C 1 3 2 6.50 3 10 2 9 C 42 5 2 737 V W B S A 5 q 1 V B 2 V A 2 . V 5 kq r . V 5 1 0.440 m 0.240 m 2 1 71.5 V 2 5 131 V. r 5 0.240 m, V 5 k q r 5 1 8.99 3 10 9 N # m 2 / C 2 21 3.50 3 10 2 9 C 2 0.200 m 1 0.240 m 5 71.5 V V 2 5 k q r 2 5 k q 3 r 1 5 V / 3 V 1 5 k q r 1 . r 2 5 3 r 1 . V 5 k q r . Electric Potential and Capacitance 18-5

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18.25. Set Up: Apply with and Solve: (a) An electron gains energy when it moves to higher potential, so and so (b) Reflect: A positive charge gains kinetic energy when it moves from high potential to low potential but a negative charge gains kinetic energy when it moves from low potential to high potential. 18.26. Set Up: From Problem 18.25, for a particle with charge of magnitude e. The speed of light is An electron has mass and a proton has mass Solve: (a) (b) From Problem 18.25, the kinetic energy the particle gains is so gives electrons and protons the same kinetic energy. But the protons must be accelerated by a potential decrease whereas the electrons are acceler- ated by a potential increase. (c) 18.27. Set Up: and Solve: (a) so which is constant. Let the accelerating voltage. Then and Thus (b) so Reflect: If the charge is positive it is accelerated by a decrease in potential. If it is negative it is accelerated by an increase in potential. 18.28. Set Up: Example 18.4 shows that where y is the distance from the negative plate. at the negative plate. For The potential gradient is Solve: (a) when The equipotential surface is a flat sheet parallel to the plates and midway between them. (b) The V equipotential surface is a flat sheet parallel to the plates and 1.0 cm from the negative plate and 5.0 cm from the positive plate. (c) D V D y 5 12.0 V 6.0 3 10 2 2 m 5 200 V / m 1 2.0 y 5 V E 5 2.0 V 2.0 V / cm 5 1.0 cm. 1 6.0 V y 5 3.0 cm. V 5 6.0 V y 5 V E 5 6.0 V 2.0 V / cm 5 3.0 cm; E 5 12.0 V 6.0 cm 5 2.0 V / cm. D V D y . V 5 12.0 V. y 5 6.0 cm, V 5 0 V 5 Ey , V 2 5 V 1 1 v 2 2 v 1 2 2 5 9 V 1 5 900 V. v 2 5 3 v v 1 2 V 1 5 v 2 2 V 2 . v 5 Å V 2 V 1 v 1 5 " 2 v . v 1 5 v . V 2 5 2 V 1 v 1 " V 1 5 v 2 " V 2 . V a 2 V b 5 V , v " V a 2 V b 5 Å 2 q m , v 5 Å 2 K b m 5 Å 2 q 1 V a 2 V b 2 m K b 5 U a 2 U b 5 q 1 V a 2 V b 2 . K a 5 0 K 5 1 2 m v 2 . U 5 qV . K a 1 U a 5 K b 1 U b v 5 Å 2 eV m 5 Å 2 1 1.60 3 10 2 19 C 21 26 V 2 1.67 3 10 2 27 kg 5 7.0 3 10 4 m / s 5 1 0.024% 2 c V 5 26 V K 5 eV V 5 m v 2 2 e 5 1 9.11 3 10 2 31 kg 21 3.00 3 10 6 m / s 2 2 2 1 1.60 3 10 2 19 C 2 5 26 V v 5 1 0.010 2 c 5 3.00 3 10 6 m / s. 1.67 3 10 2 27 kg. 9.11 3 10 2 31 kg c 5 3.00 3 10 8 m / s. v 5 " 2 eV / m v 5 Å 2 1 1.60 3 10 2 19 C 21 95 V 2 9.11 3 10 2 31 kg 5 5.8 3 10 6 m / s v 5 Å 2 K b m 5 Å 2 eV m .
• Fall '09
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