hw5s - Problem 3.45 Vector eld E is characterized by the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 3.45 Vector field E is characterized by the following properties: (a) E points along ˆ R , (b) the magnitude of E is a function of only the distance from the origin, (c) E vanishes at the origin, and (d) · E = 12, everywhere. Find an expression for E that satisfies these properties. Solution: According to properties (a) and (b), E must have the form E = ˆ R E R where E R is a function of R only. · E = 1 R 2 R ( R 2 E R ) = 12 , R ( R 2 E R ) = 12 R 2 , i R 0 R ( R 2 E R ) dR = i R 0 12 R 2 , R 2 E R | R 0 = 12 R 3 3 v v v v R 0 , R 2 E R = 4 R 3 . Hence, E R = 4 R , and E = ˆ R 4 R .
Background image of page 1

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

View Full Document Right Arrow Icon
Problem 3.47 For the vector field E = ˆ r 10 e r ˆ z 3 z , verify the divergence theorem for the cylindrical region enclosed by r = 2, z = 0, and z = 4. Solution: a i E · d s = i 2 r = 0 i 2 π φ = 0 ( ( ˆr 10 e r ˆz 3 z ) · ( r dr d ) )v v z = 0 + i 2 = 0 i 4 z = 0 ( ( 10 e r 3 z ) · ( r d dz ) )v v r = 2 + i 2 r = 0 i 2 = 0 ( ( 10 e r 3 z ) · ( r dr d ) )v v z = 4 = 0 + i 2 = 0 i 4 z = 0 10 e 2 2 d + i 2 r = 0 i 2 = 0 12 r dr d = 160 e 2 48 ≈ − 82 . 77 , iii · E d V = i 4 z = 0 i 2 r = 0 i 2 = 0 p 10 e r ( 1 r ) r 3 P r d dr dz = 8
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

hw5s - Problem 3.45 Vector eld E is characterized by the...

This preview shows document pages 1 - 3. Sign up to view the full document.

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