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Unformatted text preview: CHAPTER 7 7.1. Let V = 2 xy 2 z 3 and C = C . 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) v at P : This is v = D = C 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 Laplaces equation? No, since the charge density is not zero. 7.2. A potential field V exists in a region where C = f (x) . Find 2 V if v = 0. First, D = C(x) E = f (x) V . Then D = v = = ( f (x) V ) . So = ( 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 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 = 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 ( ) = 8 + df dx x = = ( as given ) Thus df/dx  x = = 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 ( , ) = 4 + C 2 . However, V ( , ) = = 4 + f ( , ) . So f ( , ) = 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 (/ 2 ) + B : a) Show that 2 V = 0: Since V is a function only of , 2 V = 1 r 2 sin ) d d sin dV d where dV d = d d A ln tan 2 (/ 2 ) + B = d d ( 2 A ln tan (/ 2 )) = A sin (/ 2 ) cos (/ 2 ) = 2 A sin Then 2 V = 1 r 2 sin ) d d sin 2 A sin = b) Select A and B so that V = 100 V and E = 500 V/m at P(r = 5 , = 60 , = 45 ) : First, E = V = 1 r dV d = 2 A r sin = 2 A 5 sin 60 = . 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 () = 1082 . 5 ln tan 2 (/ 2 ) 1089 . 3. 100 7.5. Given the potential field V = (A 4 + B 4 ) sin 4 : a) Show that 2 V = 0: In cylindrical coordinates, 2 V = 1 V + 1 2 2 V 2 = 1 ( 4 A 3 4 B 5 ) sin 4 1 2 16 (A 4 + B 4 ) sin 4 = 16 (A 3 + B 5 ) sin 4 16 2 (A 4 + B 4 ) sin 4 = b) Select A and B so that V = 100 V and  E  = 500 V/m at P( = 1 , = 22 . 5 , z = 2 ) : First, E = V = V a 1 V a = 4 h (A 3 B 5 ) sin 4...
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 Spring '08
 Brown
 Electromagnet

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