Lecture 34 • Electric potential: inside capacitor; of point charge; of many charges • chapter 30 (Potential and Field) Electric Potential inside a Parallel Plate Capacitor E = η due to source charges on plates • "0 U = U = qEs elec q+sources ⇒ ∆Vc potential difference: ∆Vc = V+ V = Ed; E = ( 1 N/C = 1 V/m ) • − − d V = ∆VC (d x) (decreases from positive to negative plate) d − • electric field vectors to (imaginary) equipotential surfaces/ contour lines; potential⊥ decreases along direction of E choice of zero of potential (V + or V = 0 or... ): no physical difference • − Electric Potential of Point Charge; Charged Sphere qq! Uq+q = ! 4π"0r ⇒ (scalar vs. vector E 1/r2) ∝ • 4 graphical representations • Outside sphere (same as point charge at center):V = 1 Q ; r R 4π"0 r ≥ • In terms of potential at surface, V = V (at r = R)= Q : 0 4π"0R R Q =4π"0RV0 and V = r V0 Electric Potential of Many Charges Principle of superposition (like for E): V = 1 qi • i 4π"0 ri • Continuous distribution of charge (like for E,! easier due to scalar) model as simple shape, uniform charge distribution draw picture...identify P where V to be calculated... divide Q into ∆ Q (shapes for which V known) V= i Vi ∆Q charge density dx Sum→! integral × → Potential of Ring of Charge ri same for all charge segments N N 1 ∆Q 1 1 N V = Vi = = ∆Q i=1 i=1 4π"0 ri 4π"0 √R2+z2 i=1 No need for integral: N ∆Q = Q ! ! i=1 ⇒ ! V = 1 Q ring on axis 4π"0!√R2+z2 Check: z R Vring Q/ (4π"0z) (point charge," as⇒ expected)≈ Potential of Disk of Charge uniform surface charge density η = Q/A = Q/ πR2 Use potential of ring V = 1 ∆Qi i 4π"0 2 2 √ri +z ! " V = V = 1 N ∆Q i=1 i 4π"0 i=1 2 2 √ri +z ∆Qi #= η∆Ai; ∆Ai =2#πri∆r 1 N 2Q ri∆ri Q R rdr V = 2 2 4π"0 i=1 R 2 2 2π"0R 0 √r2+z2 √ri +z → 2 2 Q 2 2 Charge of# variables: u = r + z ... V$ = 2 √R + z z disk on axis 2π"0R − ! " Chapter 30 (Potential and Field) • calculate V from E and vice versa (E and V not independent, two different mathematical representations of how source charges alter space around them); geometry of E and V • sources of potential (batteries); capacitors; currents in wires chapter 31 (Electric Circuits) • Finding Potential from Electric Field Using (i) V = Uq+sources/q; sf (ii) ∆U = W (i f)= Fsds and − → − si (iii) F¯ = qE¯ ⇒ ! Uniform E: ∆V = E ∆s • − s • Choose zero point of potential to assign V (often at ) ∞ Finding Potential from Electric Field • Point Charge ∞ ∆V = V ( ) V (r)= r Erdr ∞ 1− q − with Er = 2 and V ( ) = 0 4π"0 r ! q dr 1 ∞q ⇒ V (r)= ∞ 2 = (as before) 4π"0 r r 4π"0 r ! • Disk of Charge Q z E = 2 1 1/2 2πR "0 − (z2+R2) with V ( )! = 0 " ∞∞ ⇒ V (z)= z Ez(z)dz Q z = 2 ∞ 1 1/2 dz 2πR "0# z − (z2+R2) 2 2 Charge of# variables;! u = z +" R ... Q 2 2 V = 2 √z + R z disk 2πR "0 (as before) − $ %.