16. 3 The distribution of electric field and calculation @the perpendicular bisector of the dipole E=E+E Qi or d 研F4mr O 4 60y 4 P E B prEy 816.3 The distribution of electric field and calculation 2. The electric dipole in the electric field F DE +2 2=-,xF2=-,x(-QE 0-Q T=1+T ×QE+×QE =Qd×E=p×E
11 3 0 3 0 3 0 3 0 3 0 4 4 ( ) 4 ) 4 ( 4 Q y p y Qd r r y Q r Qr r r E E E πε πε πε πε πε r r r r r r r r r = − ≈ − = − = + − = + + − − − + + + − 2the perpendicular bisector of the dipole §16.3 The distribution of electric field and calculation − Q + Q B y +r − r r r o z y E r E+ r E− r d r 2. The electric dipole in the electric field QE d F d r r r r r = × = × 2 2 1 1 τ ( ) 2 2 2 2 QE d F d r r r r r τ = − × = − × − Qd E p E QE d QE d r r r r r r r r r r r = × = × = × + × = + 2 2 1 2 τ τ τ + Q − Q §16.3 The distribution of electric field and calculation
16. 3 The distribution of electric field and calculation 3. The electric field of continuous distributions of cha the charge distribution is imagined to be composed of a continuous sea of pointlike charges @the differential bit of charge do produces a differential bit of electric field dE dE a dl do de= dg=ods 4 pdv 816.3 The distribution of electric field and calculation Othe total electric field is the vector sum of all the contributions from all bits of charge E= de charge 4 Electric field in space depends on > The charge and its sign The geometric shape of the charge distribution >The distance from the charge distribution >The displacement of the point with respect to the charge distribution
12 3. The electric field of continuous distributions of charge 1the charge distribution is imagined to be composed of a continuous sea of pointlike charges. 2the differential bit of charge dQ produces a differential bit of electric field E r d ⎪ ⎩ ⎪ ⎨ ⎧ = V S l Q d d d d ρ σ λ r r Q E ˆ 4 d d 2 πε 0 = r E r d dQ r r P §16.3 The distribution of electric field and calculation 3the total electric field is the vector sum of all the contributions from all bits of charge. r r Q E ˆ d 4 1 dist. charge 2 0 ∫ = πε r Electric field in space depends on: ¾The charge and its sign ¾The geometric shape of the charge distribution ¾The distance from the charge distribution ¾The displacement of the point with respect to the charge distribution §16.3 The distribution of electric field and calculation
16. 3 The distribution of electric field and calculation Example 1: A ring of radius r has a total charge Q smeared out uniformly along its circumference. Calculate the electric field of the ring at a point p along the axis of the ring a distance z from the center of the ring. eiRP 816.3 The distribution of electric field and calculation Solution:dO=λds= ds 2元R do do de= 4丌Enr dE ads R 4兀Enr de E dE1=0 do
13 Example 1: A ring of radius R has a total charge Q smeared out uniformly along its circumference. Calculate the electric field of the ring at a point P along the axis of the ring a distance z from the center of the ring. R o P z z Q §16.3 The distribution of electric field and calculation s R Q Q s d 2 d d π = λ = r r s r r Q E ˆ 4 d ˆ 4 d d 2 0 2 0 πε λ πε = = r = d = 0 E⊥ ∫ E⊥ Solution: r r dQ o E r d θ θ R P z z dQ′ E′ r d r′ r §16.3 The distribution of electric field and calculation
816.3 The distribution of electric field and calculation E 1 ods ∥ cos ek survy 4丌Enr22mR 0z 4丌Enr2r2mR 4mE60(z2+R At center of the ring E=0 z→∞E→0 E z>>R E Q R 4兀E0 de R rom =0 then z=士 2 E=E max 816.3 The distribution of electric field and calculation dE Example 2: Find the electric field at a distance z along the axis of a uniformly charged circular disk of radius R and charge o per unit area Solution: d0=? dE=? e=dE=? de zdo 4mn(z2+r2) do=rro dr
14 k ˆ 4 ( ) k ˆ d 2 1 4 k ˆ cos 2 d 4 1 2 3 2 2 0 2 0 2 0 2 0 surviv // z R Qz s r R z r Q R Q s r E E R + = ⋅ = = = ⋅ ⋅ ∫ ∫ πε πε π θ πε π π r r max 2 0 then d d from E E R z z E = = = ± E = 0 z → ∞ E → 0 2 0 4 z Q z R E πε >> ≈ At center of the ring 2 R − 2 R O E z §16.3 The distribution of electric field and calculation ∫ = = = = d ? d ? d ? E E Q E dQ = 2πrσ dr 2 3 2 2 0 4 ( ) d d z r z Q E + ⋅ = πε Example 2: Find the electric field at a distance z along the axis of a uniformly charged circular disk of radius R and charge σ per unit area. Solution: §16.3 The distribution of electric field and calculation
816.3 The distribution of electric field and calculation ·z·2丌rdr E 04E0(x2+r 2o1 I/2 Jk 478 0 (z+R2) 2c[1 jk 48 (z2+R2) R→∞ E= lim 2 k 48 (z2+R2) 2 816.3 The distribution of electric field and calculation
15 k z R z k z R z k z r z r r E ˆ ] ( ) 2 [1 4 1 ˆ ] ( ) 2 [1 4 1 ˆ 4 ( ) 2 d 2 2 1 / 2 0 2 2 1 / 2 0 0 2 3 2 2 0 + = − + = − + ⋅ ⋅ = ∫ ∞ σ ε πσ πε πε r σ π k k z R z E R R ˆ 2 ˆ ] ( ) 2 [1 4 1 lim 0 2 2 1 / 2 0 ε σ σ ε = + = − → ∞ →∞ r §16.3 The distribution of electric field and calculation §16.3 The distribution of electric field and calculation