20075ee1_1_HW4

20075ee1_1_HW4 - CHAPTER 4 4.1. The value of E at P ( ρ =...

Info iconThis preview shows pages 1–3. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 4 4.1. The value of E at P ( ρ = 2 , φ = 40 ◦ , z = 3) is given as E = 100 a ρ − 200 a φ + 300 a z V/m. Determine the incremental work required to move a 20 µ C charge a distance of 6 µ m: a) in the direction of a ρ : The incremental work is given by dW = − q E · d L , where in this case, d L = dρ a ρ = 6 × 10 − 6 a ρ . Thus dW = − (20 × 10 − 6 C)(100V / m)(6 × 10 − 6 m) = − 12 × 10 − 9 J = − 12nJ b) in the direction of a φ : In this case d L = 2 dφ a φ = 6 × 10 − 6 a φ , and so dW = − (20 × 10 − 6 )( − 200)(6 × 10 − 6 ) = 2 . 4 × 10 − 8 J = 24 nJ c) in the direction of a z : Here, d L = dz a z = 6 × 10 − 6 a z , and so dW = − (20 × 10 − 6 )(300)(6 × 10 − 6 ) = − 3 . 6 × 10 − 8 J = − 36 nJ d) in the direction of E : Here, d L = 6 × 10 − 6 a E , where a E = 100 a ρ − 200 a φ + 300 a z [100 2 + 200 2 + 300 2 ] 1 / 2 = 0 . 267 a ρ − . 535 a φ + 0 . 802 a z Thus dW = − (20 × 10 − 6 )[100 a ρ − 200 a φ + 300 a z ] · [0 . 267 a ρ − . 535 a φ + 0 . 802 a z ](6 × 10 − 6 ) = − 44 . 9nJ e) In the direction of G = 2 a x − 3 a y + 4 a z : In this case, d L = 6 × 10 − 6 a G , where a G = 2 a x − 3 a y + 4 a z [2 2 + 3 2 + 4 2 ] 1 / 2 = 0 . 371 a x − . 557 a y + 0 . 743 a z So now dW = − (20 × 10 − 6 )[100 a ρ − 200 a φ + 300 a z ] · [0 . 371 a x − . 557 a y + 0 . 743 a z ](6 × 10 − 6 ) = − (20 × 10 − 6 )[37 . 1( a ρ · a x ) − 55 . 7( a ρ · a y ) − 74 . 2( a φ · a x ) + 111 . 4( a φ · a y ) + 222 . 9](6 × 10 − 6 ) where, at P , ( a ρ · a x ) = ( a φ · a y ) = cos(40 ◦ ) = 0 . 766, ( a ρ · a y ) = sin(40 ◦ ) = 0 . 643, and ( a φ · a x ) = − sin(40 ◦ ) = − . 643. Substituting these results in dW = − (20 × 10 − 6 )[28 . 4 − 35 . 8 + 47 . 7 + 85 . 3 + 222 . 9](6 × 10 − 6 ) = − 41 . 8nJ 42 4.2. An electric field is given as E = − 10 e y (sin2 z a x + x sin2 z a y + 2 x cos2 z a z ) V/m. a) Find E at P (5 , , π/ 12): Substituting this point into the given field produces E P = − 10[sin( π/ 6) a x + 5sin( π/ 6) a y + 10cos( π/ 6) a z ] = − h 5 a x + 25 a y + 50 √ 3 a z i b) How much work is done in moving a charge of 2 nC an incremental distance of 1 mm...
View Full Document

This homework help was uploaded on 04/01/2008 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.

Page1 / 5

20075ee1_1_HW4 - CHAPTER 4 4.1. The value of E at P ( ρ =...

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