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chap7 hayt

chap7 hayt - Chapter 7 Odd-Numbered 7.1 Let V = 2xy 2 z 3...

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Chapter 7 — Odd-Numbered 7.1. Let V = 2 xy 2 z 3 and = 0 . Given point P (1 , 2 , 1), find: a) V at P : Substituting the coordinates into V , find V P = 8V . b) E at P : We use E = −∇ V = 2 y 2 z 3 a x 4 xyz 3 a y 6 xy 2 z 2 a z , which, when evaluated at P , becomes E P = 8 a x + 8 a y 24 a z V / m c) ρ v at P : This is ρ v = ∇ · D = 0 2 V = 4 xz ( z 2 + 3 y 2 ) C / m 3 d) the equation of the equipotential surface passing through P : At P , we know V = 8V, so the equation will be xy 2 z 3 = 4 . e) the equation of the streamline passing through P : First, E y E x = dy dx = 4 xyz 3 2 y 2 z 3 = 2 x y Thus ydy = 2 xdx, and so 1 2 y 2 = x 2 + C 1 Evaluating at P , we find C 1 = 1. Next, E z E x = dz dx = 6 xy 2 z 2 2 y 2 z 3 = 3 x z Thus 3 xdx = zdz, and so 3 2 x 2 = 1 2 z 2 + C 2 Evaluating at P , we find C 2 = 1. The streamline is now specified by the equations: y 2 2 x 2 = 2 and 3 x 2 z 2 = 2 f) Does V satisfy Laplace’s equation? No, since the charge density is not zero. 7.3. Let V ( x, y ) = 4 e 2 x + f ( x ) 3 y 2 in a region of free space where ρ v = 0. It is known that both E x and V are zero at the origin. Find f ( x ) and V ( x, y ): Since ρ v = 0, we know that 2 V = 0, and so 2 V = 2 V ∂x 2 + 2 V ∂y 2 = 16 e 2 x + d 2 f dx 2 6 = 0 Therefore d 2 f dx 2 = 16 e 2 x + 6 df dx = 8 e 2 x + 6 x + C 1 Now E x = ∂V ∂x = 8 e 2 x + df dx and at the origin, this becomes E x (0) = 8 + df dx x =0 = 0(as given) 73
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7.3. (continued) Thus df/dx | x =0 = 8, and so it follows that C 1 = 0. Integrating again, we find f ( x, y ) = 4 e 2 x + 3 x 2 + C 2 which at the origin becomes f (0 , 0) = 4 + C 2 . However, V (0 , 0) = 0 = 4 + f (0 , 0). So f (0 , 0) = 4 and C 2 = 0. Finally, f ( x, y ) = 4 e 2 x + 3 x 2 , and V ( x, y ) = 4 e 2 x 4 e 2 x + 3 x 2 3 y 2 = 3( x 2 y 2 ) . 7.5. Given the potential field V = ( 4 + 4 )sin4 φ : a) Show that 2 V = 0: In cylindrical coordinates, 2 V = 1 ρ ∂ρ ρ ∂V ∂ρ + 1 ρ 2 2 V ∂φ 2 = 1 ρ ∂ρ ( ρ (4 3 4 5 ) ) sin4 φ 1 ρ 2 16( 4 + 4 )sin4 φ = 16 ρ ( 3 + 5 )sin4 φ 16 ρ 2 ( 4 + 4 )sin4 φ = 0 b) Select A and B so that V = 100 V and | E | = 500 V/m at P ( ρ = 1 , φ = 22 . 5 , z = 2): First, E = −∇ V = ∂V ∂ρ a ρ 1 ρ ∂V ∂φ a φ = 4 ( 3 5 )sin4 φ a ρ + ( 3 + 5 )cos4 φ a φ and at P , E P = 4( A B ) a ρ . Thus | E P | = ± 4( A B ). Also, V P = A + B . Our two equations are: 4( A B ) = ± 500 and A + B = 100 We thus have two pairs of values for A and B : A = 112 . 5 , B = 12 . 5 or A = 12 . 5 , B = 112 . 5 7.7. Let V = (cos2 φ ) in free space. a) Find the volume charge density at point A (0 . 5 , 60 , 1): Use Poisson’s equation: ρ v = 0 2 V = 0 1 ρ ∂ρ ρ ∂V ∂ρ + 1 ρ 2 2 V ∂φ 2 = 0 1 ρ ∂ρ cos2 φ ρ 4 ρ 2 cos2 φ ρ = 3 0 cos2 φ ρ 3 So at A we find: ρ vA = 3 0 cos(120 ) 0 . 5 3 = 12 0 = 106pC / m 3 b) Find the surface charge density on a conductor surface passing through B (2 , 30 , 1): First, we find E : E = −∇ V = ∂V ∂ρ a ρ 1 ρ ∂V ∂φ a φ = cos2 φ ρ 2 a ρ + 2sin2 φ ρ 2 a φ 74
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7.7b. (continued) At point B the field becomes E B = cos60 4 a ρ + 2sin60 4 a φ = 0 . 125 a ρ + 0 . 433 a φ The surface charge density will now be ρ sB =
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