Hw3 - ECE3025A Problem Set 3 Solutions 4.24 The surface dened by the equation x3 y 2 z = 1000 where x y and z are positive is an equipotential

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ECE3025A Problem Set 3 Solutions 4.24. The surface deFned by the equation x 3 + y 2 + z = 1000, where x , y , and z are positive, is an equipotential surface on which the potential is 200 V. If | E | = 50 V/m at the point P (7 , 25 , 32) on the surface, Fnd E there: ±irst, the potential function will be of the form V ( x, y, z )= C 1 ( x 3 + y 2 + z )+ C 2 , where C 1 and C 2 are constants to be determined ( C 2 is in fact irrelevant for our purposes). The electric Feld is now E = −∇ V = C 1 (3 x 2 a x +2 y a y + a z ) And the magnitude of E is | E | = C 1 ! 9 x 4 +4 y 2 + 1, which at the given point will be | E | P = C 1 ! 9(7) 4 + 4(25) 2 + 1 = 155 . 27 C 1 =50 C 1 =0 . 322 Now substitute C 1 and the given point into the expression for E to obtain E P = (47 . 34 a x +16 . 10 a y +0 . 32 a z ) The other constant, C 2 , is needed to assure a potential of 200 V at the given point. 4.32. Using Eq. (36), a) Fnd the energy stored in the dipole Feld in the region r>a : We start with E ( r, θ )= qd 4 π± 0 r 3 [2 cos θ a r + sin θ a θ ] Then the energy will be W e = " vol 1 2 ± 0 E · E dv = " 2 π 0 " π 0 " a ( qd ) 2 32 π 2 ± 0 r 6 # 4 cos 2 θ + sin 2 θ $ % ( 3 cos 2 θ +1 r 2 sin θdrdθdφ = 2 π ( qd ) 2 32 π 2 ± 0 1 3 r 3 ± ± ± a " π 0 # 3 cos 2 θ +1 $ sin θdθ = ( qd ) 2 48 π 2 ± 0 a 3 # cos 3 θ cos θ $ π 0 % ( 4 = ( qd ) 2 12 π± 0 a 3 J b) Why can we not let a approach zero as a limit? ±rom the above result, a singularity in the energy occurs as a 0. More importantly, a
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This note was uploaded on 10/28/2008 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.

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Hw3 - ECE3025A Problem Set 3 Solutions 4.24 The surface dened by the equation x3 y 2 z = 1000 where x y and z are positive is an equipotential

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