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Unformatted text preview: CHAPTER 7 7.1. Let V = 2 xy 2 z 3 and ² = ² . 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 = − ² ∇ 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 sphericallysymmetric potential field in free space, V = V e − r/a , find: a) ρ v at r = a ; Use Poisson’s equation, ∇ 2 V = − ρ v /² , which in this case becomes − ρ v ² = 1 r 2 d dr µ r 2 dV dr ¶ = − V ar 2 d dr ³ r 2 e − r/a ´ = − V ar ³ 2 − r a ´ e − r/a from which ρ v ( r ) = ² V ar ³ 2 − r a ´ e − r/a ⇒ ρ v ( a ) = ² V 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 a e − r/a a r ⇒ E ( a ) = V a e − 1 a r V / m 1 7.2c) the total charge: The easiest way is to first find the electric ﬂux density, which from part b is D = ² E = ( ² V /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 π² V 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...
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 Fall '08
 Joshi
 Electrical Engineering, Cartesian Coordinate System, René Descartes, Magnetic Field

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