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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 d x -Q d /2 − kQd ⎛ − kQ ⎞ Ey = 2⎜ 2 = ⎟ 2 2 ⎝ x + d /4 ⎠ x + d2 /4 x2 + d 2 / 4 ( Ey ≈ − kQd x3 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 Ex = Ex ≈ +Q d x kQ ( x − d / 2) 2kQd x3 2 − kQ ( x + d / 2) xd 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 Express dq, r, sinθ in terms of θ Top, bottom give same contribution dE y = − kdq r 2 π sin θ = − 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