03. Electric Field III and Electric Flux

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1 Universit of Rhode Island PHY 204: lementar Phsics II Phsics Course Materials lectric Field III and lectric Flu Gerhard Müller Universit of Rhode Island, Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: Abstract Part three of course materials for lementar Phsics II (PHY 204), taught b Gerhard Müller at the Universit of Rhode Island. Documents will be updated periodicall as more entries become presentable. Some of the slides contain figures from the tetbook, Paul A. Tipler and Gene Mosca. Phsics for Scientists and ngineers, 5 th /6 th editions. The copright to these figures is owned b W.H. Freeman. We acknowledge permission from W.H. Freeman to use them on this course web page. The tetbook figures are not to be used or copied for an purpose outside this class without direct permission from W.H. Freeman. Recommended Citation Müller, Gerhard, "03. lectric Field III and lectric Flu" (2015). PHY 204: lementar Phsics II. Paper This Course Material is brought to ou for free and open access b the Phsics Course Materials at DigitalCommons@URI. It has been accepted for inclusion in PHY 204: lementar Phsics II b an authorized administrator of DigitalCommons@URI. For more information, please contact digitalcommons@etal.uri.edu.

2 lectric Field of Continuous Charge Distribution Divide the charge distribution into infinitesimal blocks. For 3D applications use charge per unit volume: ρ = Q/ V. For 2D applications use charge per unit area: σ = Q/ A. For 1D applications use charge per unit length: λ = Q/ L. Use Coulomb s law to calculate the electric field generated b each block. Use the superposition principle to calculate the resultant field from all blocks. Use smmetries whenever possible. z d i = k dq i ri 2 ˆr i Z d i k = X i dq r 2 ˆr dq 1 r d 1 2 r 2 d 1 dq 2 1/9/2015 [tsl30 1/19]

3 lectric Field of Charged Rod (1) Charge per unit length: λ = Q/L Charge on slice d: dq = λd d dq = λ d D L lectric field generated b slice d: d = kdq 2 = kλd 2 lectric field generated b charged rod: = kλ Z D+L D» d 2 = kλ 1 D+L D = kλ» 1 D 1 D + L = kq D(D + L) Limiting case of ver short rod (L D): kq D 2 Limiting case of ver long rod (L D): kλ D 1/9/2015 [tsl31 2/19]

4 lectric Field of Charged Rod (2) Charge per unit length: λ = Q/L Charge on slice d s : dq = λd s Trigonometric relations: p = r sin θ, s = p cot θ, s = r cos θ d s = pdθ sin 2 θ d = kλd s r 2 = kλd s 2 p sin 2 θ = kλdθ p d = d sin θ = kλ p sin θdθ = kλ p d = d cos θ = kλ p cos θdθ = kλ p Z θ2 sin θdθ = kλ (cos θ 2 cos θ 1 ) θ 1 p Z θ2 θ 1 cos θdθ = kλ p (sin θ 2 sin θ 1 ) 1/9/2015 [tsl32 3/19]

5 lectric Field of Charged Rod (3) Smmetr dictates that the resulting electric field is directed radiall. θ 2 = π θ 1, sin θ 2 = sin θ 1, cos θ 2 = cos θ 1. cos θ 1 = L/2 p L 2 /4 + R 2. R = kλ R (cos θ 2 cos θ 1 ) = kλ R L p L 2 /4 + R 2. z = kλ R (sin θ 2 sin θ 1 ) = 0. Large distance (R L): R kq R 2. Small distances (R L): R 2kλ R Rod of infinite length: 2kλ = ˆR. R 1/9/2015 [tsl33 4/19]

6 lectric Field of Charged Rod (4) Smmetr dictates that the resulting electric field is directed radiall. Charge per unit length: λ = Q/L Charge on slice d: dq = λd d = kdq r 2 = kλd d = d cos θ = = = Z +L/2 L/2 d p = kλd ( ) 3/2 = kλl p (L/2) = " kλd ( ) 3/2 kλ 2p kq p (L/2) Large distance ( L): kq 2 Small distances ( L): 2kλ # +L/2 L/2 L/2 d dq L/2 1/9/2015 [tsl33 5/19]

7 lectric Field of Charged Ring Total charge on ring: Q Charge per unit length: λ = Q/2πa Charge on arc: dq d = kdq r 2 = kdq 2 + a 2 d = d cos θ = d = k ( 2 + a 2 ) 3/2 Z 2 + a 2 = dq = kdq ( 2 + a 2 ) 3/2 kq ( 2 + a 2 ) 3/2 0 0 a : kq a 3, a : kq 2 (d /d) =0 = 0 0 = ±a/ 2 1/9/2015 [tsl34 6/19]

8 Charged Bead Moving Along Ais of Charged Ring Consider a negativel charged bead (mass m, charge q) constrained to move without friction along the ais of a positivel charged ring. Place bead on -ais near center of ring: a : kq a 3 Restoring force: F = q = k s with k s = kqq a 3 Harmonic oscillation: (t) = A cos(ωt + φ) Angular frequenc: ω = r ks m = r kqq ma /9/2015 [tsl35 7/19]

9 lectric Field of Charged Disk Charge per unit area: σ = Q πr 2 Area of ring: da = 2πada Charge on ring: dq = 2πσada R da a kdq 2πσkada d = ( 2 + a 2 = ) 3/2 ( 2 + a 2 ) 3/2 Z R» ada 1 R = 2πσk 0 ( 2 + a 2 = 2πσk ) 3/2 2 + a 2 0» = 2πσk 1 for > R 2 R : 2πσk Infinite sheet of charge produces uniform electric field perpendicular to plane. 1/9/2015 [tsl36 8/19]

10 lectric Field of Charged Rubber Band The electric field at position along the line of a charged rubber band is = kq ( + L) The value of at 1 = 1m is 1 = 16N/C. Q L = 2m 1 1 = 1m 2 = 2m = 16N/C 1 (a) What is the electric field 2 at a distance 2 = 2m from the edge of the band? (b) To what length L 2 must the band be stretched (toward the left) such that it generates the field 2 = 8N/C at 1 = 1m? 1/9/2015 [tsl37 9/19]

11 lectric Field Between Charged Rods Consider four configurations of two charged rods with equal amounts of charge per unit length λ on them P 1 2 P P P (a) Determine the direction of the electric field at points P 1, P 2, P 3, P 4. (b) Rank the electric field at the four points according to strength. 1/9/2015 [tsl38 10/19]

12 lectric Field of Charged Semicircle Consider a uniforml charged thin rod bent into a semicircle of radius R. Find the electric field generated at the origin of the coordinate sstem. Charge per unit length: λ = Q/πR Charge on slice: dq = λrdθ (assumed positive) lectric field generated b slice: d = k dq R 2 = k λ R dθ directed radiall (inward for λ > 0) Components of d : d = d cos θ, d = d sin θ lectric field from all slices added up: = kλ R Z π 0 cos θ dθ = kλ R [sin θ]π 0 = 0 Rdθ R = kλ R Z π 0 sin θ dθ = kλ R [cos θ]π 0 = 2kλ R θ 1/9/2015 [tsl329 11/19]

13 lectric Flu: Definition Consider a surface S of arbitrar shape in the presence of an electric field. Prescription for the calculation of the electric flu through S: Divide S into small tiles of area A i. Introduce vector A i = ˆn i A i perpendicular to tile. If S is open choose consistentl one of two possible directions for A i. If S is closed choose A i to be directed outward. lectric field at position of tile i: i. lectric flu through tile i: Φ () i = i A i = i A i cos θ i. lectric flu through S: Φ = P i i A i. Limit of infinitesimal tiles: Φ = R d A. lectric flu is a scalar. The SI unit of electric flu is Nm 2 /C. 1/9/2015 [tsl39 12/19]

14 lectric Flu: Illustration 1/9/2015 [tsl444 13/19]

15 lectric Flu: Application (1) Consider a rectangular sheet oriented perpendicular to the z plane as shown and positioned in a uniform electric field = (2ĵ)N/C. z 2m A 3m 4m (a) Find the area A of the sheet. (b) Find the angle between A and. (c) Find the electric flu through the sheet. 1/9/2015 [tsl40 14/19]

16 lectric Flu: Application (2) Consider a plane sheet of paper whose orientation in space is described b the area vector A = (3ĵ + 4ˆk)m 2 positioned in a region of uniform electric field = (1î + 5ĵ 2ˆk)N/C. z A (a) Find the area of the sheet. (b) Find the magnitude of the electric field. (c) Find the electric flu through the sheet. (d) Find the angle between A and. 1/9/2015 [tsl41 15/19]

17 lectric Flu: Application (3) The room shown below is positioned in an electric field = (3î + 2ĵ + 5ˆk)N/C. z 1m 2m θ (a) What is the electric flu Φ through the closed door? (b) What is the electric flu Φ through the door opened at θ = 90? (c) At what angle θ 1 is the electric flu through the door zero? (d) At what angle θ 2 is the electric flu through the door a maimum? 1/9/2015 [tsl42 16/19]

18 lectric Flu: Application (4) Consider a positive point charge Q at the center of a spherical surface of radius R. Calculate the electric flu through the surface. is directed radiall outward. Hence is parallel to d A everwhere on the surface. has the same magnitude, = kq/r 2, everwhere on the surface. The area of the spherical surface is A = 4πR 2. Hence the electric flu is Φ = A = 4πkQ. Note that Φ is independent of R. 1/9/2015 [tsl43 17/19]

19 Intermediate am I: Problem #3 (Spring 05) Consider two plane surfaces with area vectors A 1 (pointing in positive -direction) and A 2 (pointing in positive z-direction). The region is filled with a uniform electric field = (2î + 7ĵ 3ˆk)N/C. (a) Find the electric flu Φ (1) through area A 1. (b) Find the electric flu Φ (2) through area A 2. z 2m A 1 A 2 3m 4m 3m 1/9/2015 [tsl333 18/19]

20 Intermediate am I: Problem #3 (Spring 05) Consider two plane surfaces with area vectors A 1 (pointing in positive -direction) and A 2 (pointing in positive z-direction). The region is filled with a uniform electric field = (2î + 7ĵ 3ˆk)N/C. (a) Find the electric flu Φ (1) through area A 1. (b) Find the electric flu Φ (2) through area A 2. z 2m 3m Solution: (a) A1 = 6î m 2, Φ (1) = A 1 = (2N/C)(6m 2 ) = 12Nm 2 /C. A 1 A 2 4m 3m 1/9/2015 [tsl333 18/19]

21 Intermediate am I: Problem #3 (Spring 05) Consider two plane surfaces with area vectors A 1 (pointing in positive -direction) and A 2 (pointing in positive z-direction). The region is filled with a uniform electric field = (2î + 7ĵ 3ˆk)N/C. (a) Find the electric flu Φ (1) through area A 1. (b) Find the electric flu Φ (2) through area A 2. z 2m 3m Solution: (a) A1 = 6î m 2, Φ (1) (b) A2 = 12ˆk m 2, Φ (2) = A 1 = (2N/C)(6m 2 ) = 12Nm 2 /C. = A 2 = ( 3N/C)(12m 2 ) = 36Nm 2 /C. A 1 A 2 4m 3m 1/9/2015 [tsl333 18/19]

22 Unit am I: Problem #2 (Spring 13) Consider three plane surfaces (one circle and two rectangles) with area vectors A 1 (pointing in positive -direction), A 2 (pointing in negative z-direction), and A 3 (pointing in negative -direction) as shown. The region is filled with a uniform electric field = ( 3î + 9ĵ 4ˆk)N/C or = (2î 6ĵ + 5ˆk)N/C. (a) Find the electric flu Φ (1) (b) Find the electric flu Φ (2) through surface 1. through surface 2. z (c) Find the electric flu Φ (3) through surface 3. 4m 3m A 3 A 1 3m 4m 3m A 2 1/9/2015 [tsl449 19/19]

23 Unit am I: Problem #2 (Spring 13) Solution: (a) A1 = π(1.5m) 2 î = 7.07m 2 î, Φ (1) A 1 = π(1.5m) 2 î = 7.07m 2 î, Φ (1) = A 1 = ( 3N/C)(7.07m 2 ) = 21.2Nm 2 /C. = A 1 = (2N/C)(7.07m 2 ) = 14.1Nm 2 /C. (b) A2 = (3m)(4m)( ˆk) = 12m 2ˆk, Φ (2) A 2 = (3m)(4m)( ˆk) = 12m 2ˆk, Φ (2) = A 2 = ( 4N/C)( 12m 2 ) = 48Nm 2 /C. = A 2 = (5N/C)( 12m 2 ) = 60Nm 2 /C. (b) A 3 = (3m)(4m)( ĵ) = 12m 2 ĵ, Φ (3) A 3 = (3m)(4m)( ĵ) = 12m 2 ĵ, Φ (3) = A 3 = (9N/C)( 12m 2 ) = 108Nm 2 /C. = A 3 = ( 6N/C)( 12m 2 ) = 72Nm 2 /C. 1/9/2015 [tsl449 19/19]