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CHAPTER 7 7.1. Let V =2xy 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 = 2y 2 z 3 a x 4xyz 3 a y 6xy 2 z 2 a z , which, when evaluated at P , becomes E P =8a x +8a y 24a z V/m c) ρ v at P : This is ρ v = ∇· D = 0 2 V = 4xz (z 2 +3y 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 = 4xyz 3 2y 2 z 3 = 2x y Thus ydy =2xdx, 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 = 6xy 2 z 2 2y 2 z 3 = 3x z Thus 3xdx = 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 2x 2 =2 and 3x 2 z 2 =2 f) Does V satisfy Laplace’s equation? No, since the charge density is not zero. 7.2. Given the spherically-symmetric potential ﬁeld in free space, V = V 0 e r/a , ﬁnd: a) ρ v at r = a; Use Poisson’s equation, 2 V = ρ v /, which in this case becomes ρ v 0 = 1 r 2 d dr r 2 dV dr = V 0 ar 2 d dr r 2 e r/a = V 0 ar 2 r a e r/a from which ρ v (r)= 0 V 0 ar 2 r a e r/a ρ v (a)= 0 V 0 a 2 e 1 C/m 3 b) the electric ﬁeld at r = a; this we ﬁnd through the negative gradient: E(r)= −∇V = dV dr a r = V 0 a e r/a a r E(a)= V 0 a e 1 a r V/m 1
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Solucionario TEORIA ELECTROMAGNETICA 7ª EDICION - Hayt Jr. & Buck

### Transcript of Capitulo 7, 7ma edición

• 1.CHAPTER 7 7.1. Let V = 2xy2 z3 and = 0. Given point P(1, 2, 1), nd: a) V at P: Substituting the coordinates into V , nd VP = 8 V. b) E at P: We use E = V = 2y2 z3 ax 4xyz3 ay 6xy2 z2 az, which, when evaluated at P, becomes EP = 8ax + 8ay 24az V/m c) v at P: This is v = D = 02 V = 4xz(z2 + 3y2 ) C/m3 d) the equation of the equipotential surface passing through P: At P, we know V = 8 V, so the equation will be xy2 z3 = 4. e) the equation of the streamline passing through P: First, Ey Ex = dy dx = 4xyz3 2y2z3 = 2x y Thus ydy = 2xdx, and so 1 2 y2 = x2 + C1 Evaluating at P, we nd C1 = 1. Next, Ez Ex = dz dx = 6xy2 z2 2y2z3 = 3x z Thus 3xdx = zdz, and so 3 2 x2 = 1 2 z2 + C2 Evaluating at P, we nd C2 = 1. The streamline is now specied by the equations: y2 2x2 = 2 and 3x2 z2 = 2 f) Does V satisfy Laplaces equation? No, since the charge density is not zero. 7.2. Given the spherically-symmetric potential eld in free space, V = V0er/a , nd: a) v at r = a; Use Poissons equation, 2 V = v/ , which in this case becomes v 0 = 1 r2 d dr r2 dV dr = V0 ar2 d dr r2 er/a = V0 ar 2 r a er/a from which v(r) = 0V0 ar 2 r a er/a v(a) = 0V0 a2 e1 C/m3 b) the electric eld at r = a; this we nd through the negative gradient: E(r) = V = dV dr ar = V0 a er/a ar E(a) = V0 a e1 ar V/m 1