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chap7 - CHAPTER 7 www.usmanahmad.cjb.net 7.1 Let V = 2xy 2...

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CHAPTER 7 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 = − 8 V . 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 = − 8 V, 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. 7.2. A potential field V exists in a region where = f (x) . Find 2 V if ρ v = 0. First, D = (x) E = − f (x) V . Then ∇ · D = ρ v = 0 = ∇ · ( f (x) V ) . So 0 = ∇ · ( f (x) V ) = − df dx ∂V ∂x + f (x) 2 V ∂x 2 f (x) 2 V ∂y 2 + f (x) 2 V ∂z 2 = − df dx ∂V ∂x + f (x) 2 V Therefore, 2 V = − 1 f (x) df dx ∂V ∂x 99
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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 ) 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 ) . 7.4. Given the potential field V = A ln tan 2 (θ/ 2 ) + B : a) Show that 2 V = 0: Since V is a function only of θ , 2 V = 1 r 2 sin θ) d sin θ dV where dV = d A ln tan 2 (θ/ 2 ) + B = d ( 2 A ln tan (θ/ 2 )) = A sin (θ/ 2 ) cos (θ/ 2 ) = 2 A sin θ Then 2 V = 1 r 2 sin θ) d sin θ 2 A sin θ = 0 b) Select A and B so that V = 100 V and E θ = 500 V/m at P (r = 5 , θ = 60 , φ = 45 ) : First, E θ = −∇ V = − 1 r dV = − 2 A r sin θ = − 2 A 5 sin 60 = − 0 . 462 A = 500 So A = − 1082 . 5 V. Then V P = − ( 1082 . 5 ) ln tan 2 ( 30 ) + B = 100 B = − 1089 . 3 V Summarizing, V (θ) = − 1082 . 5 ln tan 2 (θ/ 2 ) 1089 . 3. 100
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7.5. Given the potential field V = (Aρ 4 + 4 ) sin 4 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 ) sin 4 φ 1 ρ 2 16 (Aρ 4 + 4 ) sin 4 φ = 16 ρ (Aρ 3 + 5 ) sin 4 φ 16 ρ 2 (Aρ 4 + 4 ) sin 4 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 (Aρ 3 5 ) sin 4 φ a ρ + (Aρ 3 + 5 ) cos 4 φ a φ and at P , E P = − 4 (A B) a
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