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# hw3 - ECE3025A Problem Set 3 Solutions 4.24 The surface...

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ECE3025A Problem Set 3 Solutions 4.24. The surface defined 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, find E there: First, 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 field 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) find the energy stored in the dipole field 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 θ dr dθ dφ = 2 π ( qd ) 2 32 π 2 0 1 3 r 3 a π 0 3 cos 2 θ + 1 sin θ dθ = ( 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? From the above result, a singularity in the energy occurs as a 0. More importantly, a cannot be too small, or the original far-field assumption used to derive Eq. (36) ( a >> d

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