Summary Electric potential: PE per unit charge, V=PE/q Potential difference: work done to move the charge from one point to another, Vab=Vb-Va=DPE/q=Wab/q or Wab=qVab For uniform electric field: Vab=-Ed For electric field due to a point charge: V=kQ/r, (V=0 at r=∞)

V=kQ/r (V=0 at r=∞) V V r r Electric Potential due to a Point Charge +

(a) +kQ/R (c) –1.6kQ/R (e) +4.4kQ/R (b) –2kQ/R (d) +1.6kQ/R Three point charges –Q, –Q, and +3Q are arranged along a line as shown in the sketch. What is the electric potential at the point P? (a) +kQ/R (c) –1.6kQ/R (e) +4.4kQ/R (b) –2kQ/R (d) +1.6kQ/R X

The sketch shows cross sections of equipotential surfaces between two charged conductors shown in solid black. Points on the equipotential surfaces near the conductors are labeled A, B, C, ..., H. 1. What is the magnitude of the potential difference between points A and H? (a) 100 V (b) 200 V (c) 400 V (d) 600 V (e) 700 V X 2. What is the direction of the electric field at point E? (a) toward B (b) toward G (c) toward H (d) toward C (e) toward F X 3. How much work is required to move a +6.0 µC point charge from B to F to D to A? (a) -1.2x10–3 J (b) +1.2x10–3 J (c) +3.6x10–3 J (d) –3.6x10–3 J (e) zero joules X

The sketch below shows cross section of equipotential surfaces between two charged conductors that are shown below in solid black. Various points on the equipotential surfaces near the conductors are labeled A,B,C…,I. At which of the labeled points will the electric field have the greatest magnitude? (a) G (c) A (e) D (b) I (d) H X At which of the labeled points will an electron have the greatest potential energy? (a) A (c) G (e) I (b) D (d) H X What is the potential difference between points B and E? (a) 10 V (c) 40 V (e) 60 V (b) 30 V (d) 50 V X

Capacitance Capacitor: A Device to Store Energy in the Form of an Electric Field

Capacitance A capacitor is a device used in a variety of electric circuits The capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential difference between the conductors (plates)

Capacitance, cont Units: Farad (F) 1 F = 1 C / V A Farad is very large Often will see µF or pF

Parallel-Plate Capacitor The capacitance of a device depends on the geometric arrangement of the conductors For a parallel-plate capacitor whose plates are separated by air:

Q -Q + - d + - Capacitance C=Q/V Unit: farad, F. 1 F=1 coulomb/volt

Parallel-plate Capacitor Q=AE/4πk=εoAE, where A is area of the plate, and εo is the permittivity of free space, 8.85x10-12 C2/N-m2 Or E=s (charge density)/εo V=Ed C=Q/V= εoAE/Ed=εoA/d Properties of Capacitance C: Constant for a given capacitor, independent of Q or V C depends on the geometry of the capacitor and the material between the plates.

Question: The energy content of a charged capacitor resides in its plates potential difference charge electric field Answer: d

Question: The plates of a parallel-plate capacitor of capacitance C are brought together to one-third their original separation. The capacitance is now C/9 C/3 3C 9C Answer: c

QUICK QUIZ 16.5 You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates farther apart, do the following quantities increase, decrease, or stay the same? (a) C; (b) Q; (c) E between the plates; (d) DV.

QUICK QUIZ 16.5 ANSWER (a) C decreases (b) Q stays the same (c) E stays the same (d) DV increases

Dielectric Insulating materials Capacitance increased by a factor K

Dielectric Constant, K A measure of how effective it is in reducing an electric field across the plates Q -Q Vo Co=Q/Vo Q -Q + V=Vo/K C=Q/V=KQ/Vo=KCo

QUICK QUIZ 16.6 A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Do the following quantities increase, decrease, or stay the same? (a) C; (b) Q; (c) E between the plates; (d) DV.

QUICK QUIZ 16.6 ANSWER (a) C increases (b) Q increases (c) E stays the same (d) DV remains the same

Electric Field Energy Work must be done to separate positive and negative charges against the Coulomb forces attracting them together. This work is stored as potential energy or electric field energy in a capacitor.

e The average potential difference during the charge transfer is: <V>=(Vfinal+Vinitial)/2 =(Vf+0)/2=Vf/2 The total charge transferred is Q The work done is W=U=Q<V>=(1/2) QV =(1/2) CV2=(1/2) Q2/C (C=Q/V) + - Vinitial V e - + -Q Q Vfinal V

Energy Stored in a Capacitor Energy stored = ½ Q ΔV From the definition of capacitance, this can be rewritten in different forms

C, V C, Q Q, V W, C W, V W, Q Q= CV V= Q/C 2W/Q C= Q/V W= A Summary of the various formulas for charge Q, potential difference V, capacitance and energy W Known Quantities Unknown Quantity C, V C, Q Q, V W, C W, V W, Q Q= CV (2WC)1/2 2W/V V= Q/C (2W/C)1/2 2W/Q C= Q/V 2W/V2 Q2/2W W= CV2/2 Q2/2C QV/2