701_PartUniversity Physics Solution

701_PartUniversity Physics Solution - 23-24 Chapter 23...

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23-24 Chapter 23 23.65. (a) IDENTIFY and SET UP: Problem 23.61 derived that 1 , ln( / ) ab V E bar = where a is the radius of the inner cylinder (wire) and b is the radius of the outer hollow cylinder. The potential difference between the two cylinders is . ab V Use this expression to calculate E at the specified r. EXECUTE: Midway between the wire and the cylinder wall is at a radius of 6 ( )/2 (90.0 10 m 0.140 m)/2 0.07004 m. rab =+ = × + = 3 4 6 1 50.0 10 V 9.71 10 V/m ln( / ) ln(0.140 m/90.0 10 m)(0.07004 m) ab V E × == = × × (b) IDENTIFY and SET UP: The electric force is given by Eq.(21.3). Set this equal to ten times the weight of the particle and solve for , q the magnitude of the charge on the particle. EXECUTE: 10 E Fm g = 10 qE mg = and 92 11 4 10 10(30.0 10 kg)(9.80 m/s ) 3.03 10 C mg q E × = × × EVALUATE: It requires only this modest net charge for the electric force to be much larger than the weight. 23.66. (a) IDENTIFY: Calculate the potential due to each thin ring and integrate over the disk to find the potential. V is a scalar so no components are involved. SET UP: Consider a thin ring of radius y and width dy. The ring has area 2 ydy π so the charge on the ring is (2 ). dq σ = EXECUTE: The result of Example 23.11 then says that the potential due to this thin ring at the point on the axis at a distance x from the ring is 22 00 12 44 dq dV x yx y πσ ππ ++ PP 2 2 0 0 00 0 () 222 R R Vd V x y x R x xy σσσ ⎡⎤ = + = + ⎣⎦ + ∫∫ PPP EVALUATE: For x R ± this result should reduce to the potential of a point charge with 2 . QR σπ = 2 2 1 / 2 2 2 (1 / ) / 2 ) x RxR x xRx +=+ ≈+ so 2 /2 x RxRx +− Then 0 , 22 4 4 RR Q V x xx σσ ≈= = P as expected. (b) IDENTIFY and SET UP: Use Eq.(23.19) to calculate . x E EXECUTE: 11 1. x Vx x E xR ⎛⎞ =− = ⎜⎟ ⎝⎠ EVALUATE: Our result agrees with Eq.(21.11) in Example 21.12. 23.67. (a) IDENTIFY: Use b ab a VV d −= ⋅ E l. G G SET UP: From Problem 22.48, 2 0 2 r Er R λ = P for rR (inside the cylindrical charge distribution) and 0 2 r r = P for . Let 0 V = at = (at the surface of the cylinder). EXECUTE: > Take point a to be at R and point b to be at r, where . > Let dd l= r . G G E G and d r G are both radially outward, so . dE d r ⋅= G G Thus . r Rr R E d r −= Then 0 R V = gives . r r R VE d r In this interval 0 , ( ) / 2, rRE r r λπ >= P so 0 ln . 2 rr r dr r r R λλ P EVALUATE: This expression gives 0 r V = when = and the potential decreases (becomes a negative number of larger magnitude) with increasing distance from the cylinder.
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Electric Potential 23-25 EXECUTE: rR < Take point a at r, where , < and point b at R. dE d r ⋅= Er G G as before. Thus . R r VV E d r −= Then 0 R V = gives . R r r VE d r = In this interval ( ), < 2 0 () /2 , E rr R λπ = P so 22 2 00 0 . 2 2 2 RR r R r Vd r r d r R λλ λ ππ π ⎛⎞ == = ⎜⎟ ⎝⎠ ∫∫ PP P 2 0 1. 4 r r V R =− P EVALUATE: This expression also gives 0 r V = when . = The potential is 0 /4 P at 0 r = and decreases with increasing r.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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701_PartUniversity Physics Solution - 23-24 Chapter 23...

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