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Chapter 22: Electric Field PHY2049: Chapter 22 1 Electric Field of Single Point Charge G kq E = − 2 rˆ r G kq E = 2 rˆ r PHY2049: Chapter 22 2 Example: Electric Field on Proton ÎAt surface of proton q = e = 1.6 x 10-19 C r = 10-15 m E= ÎE kq r 2 ( = )( 9 ×109 1.6 ×10−19 ( 10−15 ) 2 ) = 1.44 ×10 21 N/C points radially outward for + charge PHY2049: Chapter 22 3 E Field of Two Equal, Positive Point Charges PHY2049: Chapter 22 4 E Field of Two Equal, Unlike Point Charges PHY2049: Chapter 22 5 Field Between Two Charged Parallel Plates ÎAssume plates are much larger than separation E is approx. constant between plates E is zero outside the plates This is a capacitor! ÎE points from + plate to – plate ÎWe will calculate E in Chap. 23 Gauss’ law Proportional to surface charge density PHY2049: Chapter 22 6 1. Rank magnitude of E at P1, P2, P3. Assume charges on rings are +q and +q PHY2049: Chapter 22 7 Answer to Question #1 Î P1 has E = 0 since it is equidistant from ring A and B and they are same sign. Î P3 has largest E because it has contributions from ring A and B Î P2 has no contribution from ring B because it is at the center, thus it is only affected by ring A. ÎSo the order (smallest E to largest E) is P1, P2, P3 PHY2049: Chapter 22 8 2. Rank magnitude of E at P1, P2, P3. Assume charges on rings are +q and −q PHY2049: Chapter 22 9 Answer to Question #2 Î P1 has largest E field since it is equidistant from ring A and B and their E contributions add, rather than cancel, as in the first question. ÎHard to rank E field of P3 and P2, in my opinion. Relative distances from the two rings are different and there is a cancellation in P3. PHY2049: Chapter 22 10 Calculate E of Dipole (⊥ axis) ÎAt point x, Ex = 0 and Ey < 0. Why? +Q r θ d x r -Q Ey = r = x2 + d 2 / 4 d /2 sin θ = r −2kQ r2 d /2 − kQd ⎛ − kQ ⎞ sin θ = 2 ⎜ 2 = ⎟ 2 2 ⎝ x + d /4 ⎠ x + d2 /4 x2 + d 2 / 4 Ey ≈ − ( kp x 3 xd ) 3/ 2 p = Qd (dipole moment) PHY2049: Chapter 22 11 Calculate E of Dipole (along axis) ÎAt point x, Ex > 0 and Ey = 0. Why? −Q +Q d x x Ex = kQ ( x − d / 2) Ex ≈ 2kp x 3 2 − xd kQ ( x + d / 2) 2 = 2kQxd (x 2 −d /4 2 ) 2 p = Qd (dipole moment) PHY2049: Chapter 22 12 Finding E Field from Charge Distribution ÎPerform Each integral over charge distribution component must be calculated separately (vector addition) dE y = ÎGeneral kdq r 2 ( sin θ or cosθ ) helpful rules Use symmetry to see if any component must be zero Use symmetry to see if any component is doubled, etc. Express dq, r and trig functions in terms of “natural” variables defined by the problem Then we can integrate! PHY2049: Chapter 22 13 Center of Uniformly Charged Circle ÎE field is down. Why? Uniform distribution of charge dE y Express dq, r, sinθ in terms of θ Top, bottom give same contribution =− kdq r 2 sin θ = − π Ey = 2 × ∫ − 0 k λ rdθ r 2 k ( λ rdθ ) r 2 sin θ sin θ −2k λ π − cos θ ( )0 r −4k λ = r dq = θ q λ= πr Ey = − 4kq π r2 dq = λ ds = λ rdθ PHY2049: Chapter 22 14 Axis of Uniform Charged Ring (+Q) ÎPoint is distance z above center of charged ring, radius R charged (ind. of angle φ) Horizontal components cancel Uniformly x dEz = r 2 z sin θ = = r z θ kdq sin θ R Ez = ∫ 2π 0 Q λ= 2π R dq = λ Rdφ Ez = (z k λ zRdφ 2 +R 2 ) 3/ 2 = k ( λ Rdφ ) z +R 2 sin θ 2 z z 2 + R2 = (z kz 2πλ R 2 +R 2 ) 3/ 2 kQz ( z 2 + R2 PHY2049: Chapter 22 ) 3/ 2 Check this! 15 Behavior at Large and Small z ÎExact ÎFor ÎFor Ez = z small kQz (z +R 2 Ez → z large Ez → kQz R kQ z2 ) 2 3/ 2 3 Becomes 0 when z = 0 Coulomb’s law As expected PHY2049: Chapter 22 16 Uniformly Charged Line ÎDistance y above midpoint of charged line of length L Uniformly charged P Ex components cancel λ = Q/ L y θ L Ey = ∫ Ey = kdq r 2 y y 2 + L2 / 4 dq = λ dx L/2 L/2 k ( λ dx ) y −L / 2 y +x y +x sin θ = ∫ kλ L x 2 2 2 2 =∫ −L / 2 ky ( λ dx ) (y 2 +x ) 2 3/ 2 Check this! PHY2049: Chapter 22 17 Special Cases for Charged Line Î Exact expression Ey = kλ L y y 2 + L2 / 4 2k λ ÎInfinite line (L →∞) E y = y Î Zero length (point) E y = kλ L y2 Falls as inverse distance = kQ y2 PHY2049: Chapter 22 Coulomb’s law As expected 18 Uniformly Charged Disk ÎPoint is distance z above axis of charged disk, radius R Ez = ∫ z θ ρ σ= R Q π R2 dq = σ dA = σ ( ρ d ρ dφ ) kdq r 2 sin θ =∫ R 2π =∫ Rk ∫ k (σρ d ρ dφ ) z z2 + ρ 2 z2 + ρ 2 0 0 0 ( 2π zσρ d ρ ) (z 2 +ρ 2 ) 3/ 2 ⎛ z Ez = 2π kσ ⎜1 − ⎜ z 2 + R2 ⎝ PHY2049: Chapter 22 ⎞ ⎟⎟ ⎠ 19 Charged Disk (cont) ⎛ z ÎExact expression E z = 2π kσ ⎜ 1 − ⎜ z 2 + R2 ⎝ Îz Îz << R >> R σ Ez → 2π kσ = 2ε 0 Ez → kπ R 2σ z 2 = kQ z2 PHY2049: Chapter 22 ⎞ ⎟⎟ ⎠ Independent of z! See next chapter Coulomb’s law As expected 20