Chapter 29 Electric Potential and Field Phys 133 – Chapter 29 1 Overview Force, field, energy, potential Phys 133 – Chapter 29 2 Force, field, energy, potential or E =- dV dr ( ¶V ˆ ¶V ˆ ¶V ˆ i+ j+ k) ¶x ¶y ¶z Phys 133 – Chapter 29 3 Fields and potentials 1. a) E = 0 V/m in throughout some region of space, can you conclude that the potential V = 0 in this region? b) V = 0 V throughout some region of space. Can you conclude that the electric field E = 0 V/m in this region? Phys 133 – Chapter 29 4 Graphically convert between E and V 2. The top graph shows Ex vs. x for an electric field parallel to the x-axis. a) Draw the graph of V vs. x in this region of space. Let V = 0 at x = 0m. Add an appropriate scale on the vertical axis. (Hint: integration is the area under the curve) b) Draw a contour map above the x-axis on a diagram like the one belowright and label your equipotential lines every 20 V. 0V has been drawn already. C) Draw several electric field vectors on top of the contour map. Ex (V/m) 0V 40 20 0 2 4 1 x(m) 2 3 4 x(m) Phys 133 – Chapter 29 5 Problem: Potential of sphere --Find the electric potential everywhere for a sphere (radius R) with charge (Q) uniformly distributed. Take V=0 at infinity. --Sketch V vs r and Er vs r. From Chap 27 E field is: ì Q ïK R 3 r rˆ E =í ï K Q rˆ î r2 r<R R<r Phys 133 – Chapter 29 6 Problem: Potential of sphere (ans) Finding Vr R<r r V∞ dr For all r For r < R For R < r Vr é KQ ù = -ê 3 r2 ú ë 2R û R KQ KQ = r ¥ =- Defined V¥ = 0 KQ r KQ (R < r) r ∆V r é KQ ù = - êë r úû ¥ r Vr = r<R ∆V VR E Vr Vr definition Vr - V¥ = ì Q ïK R 3 r rˆ E =í ï K Q rˆ î r2 æ KQ 2 KQ 2ö r - 3R è 2R 3 ø 2R KQ KQ 2 From R < r : VR = = 3 R R R Vr - KQ æ KQ 2 KQ 2ö =r - 3R è 2R 3 ø R 2R 3KQ 2 KQ 2 V = R r (r < R) Phys 133 -- Chapter r 2R 3 30 2R 3 7 Problem: Potential of sphere (ans) Find the electric potential everywhere for a sphere (radius R) with charge (Q) uniformly distributed. ì 3KQ KQ 2 ï 2R - 2R 3 r (r < R) ï V =í ï Q (R < r) K ïî r ì Q ïK R 3 r Er = í Q ï K 2 î r (r < R) (R < r) Phys 133 – Chapter 29 8 Problem: Field from Potential Find the x,y and z components of the electric field, given that the electric potential of a disk is given by Vdisk Q = 2pR 2e0 ( z 2 + R2 - z ) Phys 133 – Chapter 29 9 Problem: Field from Potential (Answer) Find the z component of the electric field, given that the electric potential of a disk is given by Vdisk Q = 2pR 2e0 ( Edisk )z ( z 2 + R2 - z ) ö dV h æ z == 1ç 2 2 ÷ dz 2e 0 è z +R ø Phys 133 – Chapter 29 10 Geometry of potential/field is perp to equipotential surfaces points downhill (decreasing V) --strength proportional to spacing equipotentials Phys 133 – Chapter 29 11 Kirchhoff’s loop rule (Conservation of energy) Phys 133 – Chapter 29 12 Conductor in equilibrium: field and potential --field is zero inside conductor --field is perpendicular at surface --conductor is at equipotential (no work to move) Phys 133 – Chapter 29 13 Conductor in equilibrium: equipotentials --equipotentials are parallel to nearby conductor Phys 133 – Chapter 29 14 Do Workbook 30.4, 6, 11, & 12a Phys 133 – Chapter 29 15 Problem: Finding Potential --Find the electric potential everywhere for a point charge (q) at the center of a hollow metal sphere (inner radius a, outer radius b) with charge Q. (Take V=0 at infinity.) a --Sketch V vs r and Er vs r. b ì Kq , r<a ï 2 r ï E r (r) = í 0, a < r < b ï K(Q + q) ï , b<r î r2 Phys 133 – Chapter 29 16 Problem: Finding Potential (ans) Finding Vr ∆V Va For all r é Kq ù = - ê- ú ë r ûa r K(Q + q) K(Q + q) = r ¥ K(Q + q) Vr - V¥ = r K(Q + q) (R < r) r E Kq Kq r a K(Q + q) From before Va = Vb = b ∆V For a < r < b = Vr - K(Q + q) Kq Kq = b r a Vr = K(Q + q) æ Kq Kqö + (r < a) è r b aø Phys 133 -- Chapter 30 r Vr dr V∞ r é K(Q + q) ù = - êúû r ë ¥ Defined V¥ = 0 Vr For r < a For b < r Vr = ì Kq , r<a ï 2 r ï E r (r) = í 0, a < r < b ï K(Q + q) ï , b<r î r2 from b < r; Vb = Vr = K(Q + q) b K(Q + q) (a < r < b) b 17 Problem: finding Potential (Answer) ì q+Q K ï r ïï q+Q V =í K b ï ïK q + Q + Kqæ 1 - 1 ö ïî è r aø b b< r a< r<b r<a ì Kq , r<a ï 2 r ï E r (r) = í 0, a < r < b ï K(Q + q) ï , b<r î r2 not origin Phys 133 -- Chapter 30 18 Sources of potential: Capacitor -charge separation -not sustained Phys 133 -- Chapter 30 19 Sources of potential: Battery --sustained -- Phys 133 -- Chapter 30 20 Capacitors charge separation on capacitor Apply to capacitor E capcitor Q =(constant) e0 A Q =( )d e0 A where C = Phys 133 -- Chapter 30 e0 A d 21 Circuit geometry --generic circuit elements --imagine battery connection 1 2 1 2 parallel series Phys 133 -- Chapter 30 22 Capacitors: series equivalent Q1 = Q2 = Qequivalent Q1 Q2 Qequivalent + = C1 C2 Cequivalent 1 1 1 + = C1 C2 Cequivalent Phys 133 -- Chapter 30 23 Capacitors: parallel equivalent Q1 + Q2 = Qequivalent C1 + C2 = Cequivalent Phys 133 -- Chapter 30 24 Capacitors 1 1 2 2 series parallel V V1+V2= Veq V1=V2 Q Q1=Q2 Q1+Q2= Qeq 1 1 1 = + +… Ceq C1 C2 Ceq = C1 + C2 +… Phys 133 -- Chapter 30 25 Do Workbook 30.26 & 27 Phys 133 -- Chapter 30 26 Energy in capacitor --work done charge separation --or in electric field Phys 133 -- Chapter 30 27 Problem --Find the charge on (Q) and potential difference (V) across each capacitor. --What is the total energy stored in the system? 5 mF 2 mF 8 mF 12 V 2 mF Phys 133 -- Chapter 30 28 Cequ = C2 + C8 (parallel) Problem (ans) = 2mF +8mF 5 mF 2 mF 8 mF =10mF 12 V 2 mF 5 mF 1 1 1 1 = + + (series) Cequ C5 C10 C2 10 µF 12 V 2 mF 1 1 1 = + + 5mF 10mF 2mF 8 = 10mF 5mF Þ Cequ = 4 5/4 µF 12 V Phys 133 -- Chapter 30 29 Problem (ans) 5/4 µF 12 V Q5 = Q10 = Q2 = Qtot = 15mC (in series) 5 mF 10 µF 12 V 2 mF 5 mF 2 mF 8 mF 12 V 2 mF Phys 133 -- Chapter 30 30 Dielectrics change the potential difference • The potential between to parallel plates of a capacitor changes when the material between the plates changes. It does not matter if the plates are rolled into a tube as they are in Figure 24.13 or if they are flat as shown in Figure 24.14. C K= C0 V0 V= K E0 E= K Table 24.1—Dielectric constants Field lines as dielectrics change • Moving from part (a) to part (b) of Figure 24.15 shows the change induced by the dielectric. In vacuum, energy density is 1 u = e0 E 2 2 In dielectric e = Ke 0 C = Ke0 A A =e d d 1 2 u = eE 2 Examples to consider, capacitors with and without dielectrics • If capacitor is disconnected from circuit, inserting a dielectric changes decreases electric field, potential and increases capacitance, but the amount of charge on the capacitor is unchanged. • If the capacitor is hooked up to a power supply with constant voltage, the voltage must remain the same, but capacitance and charge increase Q24.8 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the potential difference between the capacitor plates? A. The potential difference increases. B. The potential difference remains the same. C. The potential difference decreases. D. not enough information given to decide A24.8 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the potential difference between the capacitor plates? A. The potential difference increases. B. The potential difference remains the same. C. The potential difference decreases. D. not enough information given to decide Q24.9 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide A24.9 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide Q24.10 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the amount of charge on each of the capacitor plates? A. The amount of charge increases. B. The amount of charge remains the same. C. The amount of charge decreases. D. not enough information given to decide A24.10 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the amount of charge on each of the capacitor plates? A. The amount of charge increases. B. The amount of charge remains the same. C. The amount of charge decreases. D. not enough information given to decide Q24.11 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide A24.11 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide