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Lecture 18 • Electric field due to point charge • Chapter 27 (The Electric Field) Ē due to configuration of (source) charges (today) parallel plate capacitor (uniform Ē ) motion of (other) charges in Ē • • Electric Field of Point Charge Use Coulomb’s law: ! " F̄on q! 1 q Ē = q! = 4π"0 r2 , away from q field diagram (sample vectors): at tail of vector; does not stretch... Unit Vector Notation • mathematical notation for “away from q’’ • î, ĵ, k̂ : magnitude 1 (no units), purely directional information • r̂ : unit vector pointing from origin to point (“straight outward from point” like Ē ) • applies to q < 0 (−r̂ points towards charge) Electric Field Models • good approx. for small charged object, real wire • Coulomb’s law ! K= 1 4π"0 • Ē due to point charge • limiting cases (check + simpler formula): like Ē point charge very far ( ! size) " • Electric Field of Multiple Charges Principle of superposition ! +F̄on q = F̄1 on q + F̄2 on q ⇒ Ēnet = i Ē !i Strategy: picture..., identify P, find Ēi , i Ēi (use symmetry) • Electric field of a dipole (two opposite charges separated by small distance s): no net charge, but does have a net Ē (Edipole )y = = = (E+ )y + (E− )y 1 q + 2 4π"0 (y − 1/2s) 1 −q 4π"0 (y + 1/2s)2 q 2ys 4π"0 (y − 1/2s)2 (y + 1/2s)2 Electric Field of a Dipole • For y >> s: (Edipole )y ≈ • dipole moment: p̄ = (qs, Picturing Ē • • • draw vectors or lines • • starts on +ve, end -ve tangent: direction of Ē closer together for larger Ē cannot cross (unique direction at point) 1 2qs 4π"0 y 3 from negative to positive) Electric Field of a Continuous Charge Distribution • even if charge is discrete, consider it continuous, describe how it’s distributed (like density, even if atoms • Strategy (based on of point charge and principle of superposition) divide Q into point-like charges ∆Q find Ē due to ∆Q convert sum to integral: ∆Q → density ×dx (x describes shape of ∆Q) Electric Field of a Line of Charge • • • Problem Strategy Solution...: Erod = |Q| 1 √ 4π"0 r r 2 +(L/2)2 Infinite line of charge:Eline = limL→∞ 4π"1 0 r√r2|Q| = 2 +(L/2) |Q| 1 4π"0 rL/2 = 1 2|λ| 4π"0 r Review of Line of charge • ! due to segment i: Ēi " = x = 1 ∆Q cos θi 2 4!0 ri 1 r∆Q 4!0 (r2 + y 2 )3/2 i • • • Sum over segments: !N Ex = i (Ei )x = 1 4π"0 !N r∆Q 3/2 2 2 r +yi ( ) Relate ∆Q to coordinate: ∆Q = λ∆y = (Q/L) ∆y Q/L !N r∆y Ex = 4π"0 i 3/2 2 2 (r +yi ) " L/2 !N Sum to integral: N → ∞; ∆y → dy; i → −L/2 Ex = Q/L 4π"0 ! L/2 i rdy −L/2 (r 2 +y 2 )3/2 Ring and Disk of charge • • On axis of ring: (Ering )z = Divide disk of surface ! sum: " Ering i z = zQ 1 4π"0 (z 2 +R2 )3/2 Q charge density η = A z∆Qi 1 4π"0 (z 2 +r 2 )3/2 i (Edisk )z = !N " i = Ering i # z Q πR2 = into rings; z 4π"0 !N i ∆Qi 3/2 (z2 +ri2 ) • • Disk of Charge Relate ∆Q to coordinate: ∆Qi = η∆Ai = η2πri ∆r 2πηz !N ri ∆r (Edisk )z = 4π#0 i 3/2 2 2 (z +ri ) "R !N Sum to integral: N → ∞; ∆r → dr; i → 0 (Edisk )z = • ηz 2"0 !R 0 rdr (z 2 +r 2 )3/2 For z >> R, (Edisk )z = Q 4π"0 z 2 (point charge), as expected Plane and Sphere of Charge • Electrodes (charged surfaces used to steer electrons): model as infinite plane if distance to plane, z << distance to edges • Outside sphere (or shell) of uniform charge