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CHAPTER 3 D3.1. Given a 60-μC point charge located at the origin, find the total electric flux passing through: (a) that portion of the sphere r = 26 cm bounded by 0 < 0 <л/2 and 0 < ¢ <л/2; (b) the closed surface defined by p = 26 cm and z = +26 cm; (c) the plane z = 26 cm. (a) D = a, 4π² d¥ = a,(r² sin ododo)a, = 4лr sin ǝdedo Απ y = 4π Jo Jo 4πJo 60×10 Απ ²² sinodd = d² sino deſ² do = (-cose) (²) == -(-cos 90°-(-cos 0°))((1/2)-0) (b) Deriving Gauss's law of the cylinder, 4π = 7.5 με 60×10 (-cos 90°-(-cos 0°))((1/2)-0) = 7.5 με 4π (b) Deriving Gauss's law of the cylinder, s⋅ds = Ds§ pdødz = Dspſ doſt dz = Dsp = Ds 2xpl Q=fDs d Ds = 오 2 Dp = or Dp = ap 2πρι 2πρί Solving Y, d¥ = ap (p do dz)ap = 2πρί p do dz 2πρί = dz = 2πρι - (2πρL) 2πρί y = = 60 με (c) Getting the equation in (b) with changing the limits of as 0≤≤л because only the flux lines pass in half through the plane, dy pf døft dz (лρL) 2πρι 2πρι y = 0/2 = 30 μC

CHAPTER 3 D3.1. Given a 60-μC point charge located at the origin, find the total electric flux passing through: (a) that portion of the sphere r = 26 cm bounded by 0 < 0 <л/2 and 0 < ¢ <л/2; (b) the closed surface defined by p = 26 cm and z = +26 cm; (c) the plane z = 26 cm. (a) D = a, 4π² d¥ = a,(r² sin ododo)a, = 4лr sin ǝdedo Απ y = 4π Jo Jo 4πJo 60×10 Απ ²² sinodd = d² sino deſ² do = (-cose) (²) == -(-cos 90°-(-cos 0°))((1/2)-0) (b) Deriving Gauss's law of the cylinder, 4π = 7.5 με 60×10 (-cos 90°-(-cos 0°))((1/2)-0) = 7.5 με 4π (b) Deriving Gauss's law of the cylinder, s⋅ds = Ds§ pdødz = Dspſ doſt dz = Dsp = Ds 2xpl Q=fDs d Ds = 오 2 Dp = or Dp = ap 2πρι 2πρί Solving Y, d¥ = ap (p do dz)ap = 2πρί p do dz 2πρί = dz = 2πρι - (2πρL) 2πρί y = = 60 με (c) Getting the equation in (b) with changing the limits of as 0≤≤л because only the flux lines pass in half through the plane, dy pf døft dz (лρL) 2πρι 2πρι y = 0/2 = 30 μC

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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CHAPTER 3 D3.1. Given a 60-μC point charge located at the origin, find the total electric flux passing through: (a) that portion of the sphere r = 26 cm bounded by 0 < 0 <л/2 and 0 < ¢ <л/2; (b) the closed surface defined by p = 26 cm and z = +26 cm; (c) the plane z = 26 cm. (a) D = a, 4π² d¥ = a,(r² sin ododo)a, = 4лr sin ǝdedo Απ y = 4π Jo Jo 4πJo 60×10 Απ ²² sinodd = d² sino deſ² do = (-cose) (²) == -(-cos 90°-(-cos 0°))((1/2)-0) (b) Deriving Gauss's law of the cylinder, 4π = 7.5 με 60×10 (-cos 90°-(-cos 0°))((1/2)-0) = 7.5 με 4π (b) Deriving Gauss's law of the cylinder, s⋅ds = Ds§ pdødz = Dspſ doſt dz = Dsp = Ds 2xpl Q=fDs d Ds = 오 2 Dp = or Dp = ap 2πρι 2πρί Solving Y, d¥ = ap (p do dz)ap = 2πρί p do dz 2πρί = dz = 2πρι - (2πρL) 2πρί y = = 60 με (c) Getting the equation in (b) with changing the limits of as 0≤≤л because only the flux lines pass in half through the plane, dy pf døft dz (лρL) 2πρι 2πρι y = 0/2 = 30 μC
Transcribed Image Text:CHAPTER 3 D3.1. Given a 60-μC point charge located at the origin, find the total electric flux passing through: (a) that portion of the sphere r = 26 cm bounded by 0 < 0 <л/2 and 0 < ¢ <л/2; (b) the closed surface defined by p = 26 cm and z = +26 cm; (c) the plane z = 26 cm. (a) D = a, 4π² d¥ = a,(r² sin ododo)a, = 4лr sin ǝdedo Απ y = 4π Jo Jo 4πJo 60×10 Απ ²² sinodd = d² sino deſ² do = (-cose) (²) == -(-cos 90°-(-cos 0°))((1/2)-0) (b) Deriving Gauss's law of the cylinder, 4π = 7.5 με 60×10 (-cos 90°-(-cos 0°))((1/2)-0) = 7.5 με 4π (b) Deriving Gauss's law of the cylinder, s⋅ds = Ds§ pdødz = Dspſ doſt dz = Dsp = Ds 2xpl Q=fDs d Ds = 오 2 Dp = or Dp = ap 2πρι 2πρί Solving Y, d¥ = ap (p do dz)ap = 2πρί p do dz 2πρί = dz = 2πρι - (2πρL) 2πρί y = = 60 με (c) Getting the equation in (b) with changing the limits of as 0≤≤л because only the flux lines pass in half through the plane, dy pf døft dz (лρL) 2πρι 2πρι y = 0/2 = 30 μC
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