20075ee1_1_HW7 - CHAPTER 7 7.1 Let V = 2xy 2 z 3 and = 0...

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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 = 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. 7.2. 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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7.2c) the total charge: The easiest way is to first find the electric flux density, which from part b is D = 0 E = ( 0 V 0 /a ) e r/a a r . Then the net outward flux of D through a sphere of radius r would be Φ( r ) = Q encl ( r ) = 4 πr 2 D = 4 π 0 V 0 r 2 e r/a C As r → ∞ , this result approaches zero, so the total charge is therefore Q net = 0 . 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 ( ρ, φ ) = ( V 0 ρ/d )cos φ : a) Show that V ( ρ, φ ) satisfies Laplace’s equation: 2 V = 1 ρ ∂ρ ρ ∂V ∂ρ + 1 ρ 2 2 V ∂φ 2 = 1 ρ ∂ρ V 0 ρ d cos φ 1 ρ 2 ∂φ V 0 ρ d sin φ = V 0 ρ d cos φ V 0 ρ d sin φ = 0
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