This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 7 7.1. Let V = 2 xy 2 z 3 and = . 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) v at P : This is v = D = 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 Laplaces 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) v at r = a ; Use Poissons equation, 2 V = v / , which in this case becomes v = 1 r 2 d dr r 2 dV dr = V ar 2 d dr r 2 e r/a = V ar 2 r a e r/a from which v ( r ) = V ar 2 r a e r/a v ( a ) = 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 ux density, which from part b is D = E = ( V /a ) e r/a a r . Then the net outward ux of D through a sphere of radius r would be ( r ) = Q encl ( r ) = 4 r 2 D = 4 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 v = 0. It is known that both E x and V are zero at the origin. Find f ( x ) and V ( x, y ): Since 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...
View
Full
Document
 Fall '08
 Joshi
 Electrical Engineering

Click to edit the document details