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# A de cos 1 33 b de cos 2 c de sin 1 d de sin

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Unformatted text preview: dx x 2 (B) dx 2 a +h 2 (C ) λdx 2 a +h 2 (D) λdx 2 (a − x) + h 2 (E) λdx x2 P h y s ic s 2 1 2 Le c tu re 2 , S lid e 1 6 dE Calculation C h a rg e is unifo rm ly d is trib ute d a lo n g th e x­a xis fro m th e o rig in to x = a . T h e c h a rg e d e nis ty is λ C /m . Wh a t is th e x­ c o m p o ne n t o f th e e le c tric fie ld a t p o int P : (x,y ) = (a ,h ) ? dE x r θ1 θ2 x dq ˆ E = ∫k 2 r r We kno w: dq r2 = θ2 P y λdx (a − x) 2 + h 2 h x a dq=λdx E x = ∫ dE x dE x Wh a t is ? (A) dE cos θ1 33 (B) dE cos θ 2 (C ) dE sin θ1 (D) dE sin θ 2 P h y s ic s 2 1 2 Le c tu re 2 , S lid e 1 7 dE Calculation C h a rg e is unifo rm ly d is trib ute d a lo n g th e x­a xis fro m th e o rig in to x = a . T h e c h a rg e d e nis ty is λ C /m . Wh a t is th e x­ c o m p o ne n t o f th e e le c tric fie ld a t p o int P : (x,y ) = (a ,h ) ? dq ˆ E = ∫k 2 r r θ2 P “The concept I found most difficult was integrating y The to find electric field when the charge is distributed over a distance.” over dE x r θ1 h θ2 x x a dq=λdx We kno w: dq r 2 = λdx E x = ∫ dE x = ∫ dE cos θ 2 (a − x) 2 + h 2 Ex Wh a t is ? ∞ dx (A) k λ cos θ 2 ∫ ( a − x )2 + h 2 −∞ dx ( a − x )2 + h 2 0 (B) k λ cos θ 2 ∫ (C ) neither of the above 33 a cosθ2 DEPENDS ON x !! P h y s ic s 2 1 2 Le c tu re 2 , S lid e 1 8 dE Calculation C h a rg e is unifo rm ly d is trib ute d a lo n g th e x­a xis fro m th e o rig in to x = a . T h e c h a rg e d e nis ty is λ C /m . Wh a t is th e x­ c o m p o ne n t o f th e e le c tric fie ld a t p o int P : (x,y ) = (a ,h ) ? dE x r θ1 We kno w: dq r2 = h θ2 x dq ˆ E = ∫k 2 r r θ2 P y x a dq=λdx λdx E x = ∫ dE x = ∫ dE cos θ 2 (a − x) 2 + h 2 cos θ 2 Wh a t is ? (A) 33 x 2 a +h 2 (B) a−x 2 (a − x) + h 2 (C ) a 2 a +h 2 (D) a (a − x) 2 + h 2 P h y s ic s 2 1 2 Le c tu re 2 , S lid e 1 9 dE Calculation C h a rg e is unifo rm ly d is trib ute d a lo n g th e x­a xis fro m th e o rig in to x = a . T h e c h a rg e d e nis ty is λ C /m . Wh a t is th e x­ c o m p o ne n t o f th e e le c tric fie ld a t p o int P : (x,y ) = (a ,h ) ? We kno w: dq r2 = dq ˆ E = ∫k 2 r r λdx dE x r θ1 h θ2 x x a dq=λdx E x = ∫ dE x = ∫ dE cos θ 2 (a − x) 2 + h 2 θ2 P y cos θ 2 = a−x (a − x) 2 + h 2 E x ( P) Wh a t is ? a Ex ( P ) = k λ ∫ dx 0 33 a−x ((a− x) 2 +h ) 2 3/ 2 Ex ( P ) = kλ h 1− 2 2 h h +a P h y s ic s 2 1 2 Le c tu re 2 , S lid e 2 0 dE Observation C h a rg e is unifo rm ly d is trib ute d a lo n g th e x­a xis fro m th e o rig in to x = a . T h e c h a rg e d e nis ty is λ C /m . Wh a t is th e x­ c o m p o ne n t o f th e e le c tric fie ld a t p o int P : (x,y ) = (a ,h ) ? No te th a t o ur re s ult c a n b e re writte n m o re s im p ly in te rm s o f θ1 . kλ h Ex ( P ) = 1 − 2 2 h h +a θ2 P y dE x r θ1 h θ2 x x a dq=λdx kλ Ex ( P ) = ( 1 − sin θ1 ) h Exe rc is e fo r s tud e nt: C h a ng e va ria b le s : write x in te rm s o f θ R e s ult: o b ta in s im p le inte g ra l in θ 33 kλ Ex ( P ) = h π /2 ∫ dθ cos θ θ1 P h y s ic s 2 1 2 Le c tu re 2 , S lid e 2 1 Notes • • • • • • Preflight + Prelecture 3 due by 8:00 AM Tuesday Aug. 30 Homework 1 is due Tuesday Aug. 30 Labs start Monday Aug. 29 Discussion Quiz next week will be on Coulomb’s Law and E Homework 2 is due Tuesday Sep. 6 No office hours on Monday Sep. 5 (Labor Day) Physics 212 Lecture 2, Slide 22 Physics 212 Lecture 2, Slide...
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