chap7 - CHAPTER 7 7.1. Let V = 2 xy 2 z 3 and C = C . Given...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 17

chap7 - CHAPTER 7 7.1. Let V = 2 xy 2 z 3 and C = C . Given...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online