{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# chap7 - CHAPTER 7 www.usmanahmad.cjb.net 7.1 Let V = 2xy 2...

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

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 = − 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) 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 = − 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 Laplace’s equation? No, since the charge density is not zero. 7.2. A potential field V exists in a region where = f (x) . Find 2 V if ρ v = 0. First, D = (x) E = − f (x) V . Then ∇ · D = ρ v = 0 = ∇ · ( f (x) V ) . So 0 = ∇ · ( 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

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

View Full Document
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 = A ln tan 2 (θ/ 2 ) + B : a) Show that 2 V = 0: Since V is a function only of θ , 2 V = 1 r 2 sin θ) d sin θ dV where dV = d A ln tan 2 (θ/ 2 ) + B = d ( 2 A ln tan (θ/ 2 )) = A sin (θ/ 2 ) cos (θ/ 2 ) = 2 A sin θ Then 2 V = 1 r 2 sin θ) d sin θ 2 A sin θ = 0 b) Select A and B so that V = 100 V and E θ = 500 V/m at P (r = 5 , θ = 60 , φ = 45 ) : First, E θ = −∇ V = − 1 r dV = − 2 A r sin θ = − 2 A 5 sin 60 = − 0 . 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 (θ) = − 1082 . 5 ln tan 2 (θ/ 2 ) 1089 . 3. 100
7.5. Given the potential field V = (Aρ 4 + 4 ) sin 4 φ : a) Show that 2 V = 0: In cylindrical coordinates, 2 V = 1 ρ ∂ρ ρ ∂V ∂ρ + 1 ρ 2 2 V ∂φ 2 = 1 ρ ∂ρ ρ( 4 3 4 5 ) sin 4 φ 1 ρ 2 16 (Aρ 4 + 4 ) sin 4 φ = 16 ρ (Aρ 3 + 5 ) sin 4 φ 16 ρ 2 (Aρ 4 + 4 ) sin 4 φ = 0 b) Select A and B so that V = 100 V and | E | = 500 V/m at P (ρ = 1 , φ = 22 . 5 , z = 2 ) : First, E = −∇ V = − ∂V ∂ρ a ρ 1 ρ ∂V ∂φ a φ = − 4 (Aρ 3 5 ) sin 4 φ a ρ + (Aρ 3 + 5 ) cos 4 φ a φ and at P , E P = − 4 (A B) a

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern