Chapter 15

1. Q Q F 1 2 k = 12 r 2 12 Chapter 15 • Coulomb’s Law • Electric Field • Conductors in Electric Field E=F/q, (E=E1+E2+E3+…) E=0inside a good conductor Any excess charge resides on the surface of the conductor

2. Chapter 16 Electric Potential Electric Energy

3. We have learned that • Work=force×distance • Potential energy (PE) is the energy that depends on the position or configuration • PE is a property of a system of interacting objects

4. + A m F=mg Q b F=QE - B Earth + A W=DPE W=QEd(charge) W=mgh(mass) PE=mgh (KE=0) F=mg PE=QEd (KE=0) F=QE h c d PE=0 PE=0 - B Earth + A d PE=0 KE=mgh PE=0 KE=QEd - B Earth Electric Gravitational + A g a E - B Earth

5. Work done on Charge: W=DPE=QE·d-0 with respect to B (PE=0 at B) Work done on mass: W=DPE=mg·h-0 (PE=0 on the earth) • There is no change in the energy of a charge moved perpendicular to an electric field, as there is no change in the energy of a mass moved perpendicular to a gravitational field ( for instance, along the earth’s surface)

6. Potential Difference • The potential difference between points A and B is defined as the change in the potential energy (final value minus initial value) of a charge q moved from A to B divided by the size of the charge • ΔV = VB – VA = ΔPE / q • Potential difference is not the same as potential energy

7. Potential Difference, cont. • Another way to relate the energy and the potential difference: ΔPE = q ΔV • Both electric potential energy and potential difference are scalar quantities • Units of potential difference • V = J/C • A special case occurs when there is a uniform electric field • VB – VA= -Ed • Gives more information about units: N/C = V/m

8. Potential differenceis the work done to move the charge from one point to another. Onlypotential differencebetween two pointsis physically meaningful DV=Vab=Vb-Va=Wab/Q Equipotential lines a E d b Vab=-Ed (uniform electric field)

9. Energy and Charge Movements • A positive charge gains electrical potential energy when it is moved in a direction opposite the electric field • If a charge is released in the electric field, it experiences a force and accelerates, gaining kinetic energy • As it gains kinetic energy, it loses an equal amount of electrical potential energy • A negative charge loses electrical potential energy when it moves in the direction opposite the electric field

10. Energy and Charge Movements, cont • When the electric field is directed downward, point B is at a lower potential than point A • A positive test charge that moves from A to B loses electric potential energy • It will gain the same amount of kinetic energy as it loses potential energy

11. Summary of Positive Charge Movements and Energy • When a positive charge is placed in an electric field • It moves in the direction of the field • It moves from a point of higher potential to a point of lower potential • Its electrical potential energy decreases • Its kinetic energy increases

12. Summary of Negative Charge Movements and Energy • When a negative charge is placed in an electric field • It moves opposite to the direction of the field • It moves from a point of lower potential to a point of higher potential • Its electrical potential energy decreases • Its kinetic energy increases

13. A charge q = -4.0 µC is moved 0.25 m horizontally to point P in a region where an electric field is 150 V/m and directed vertically as shown. What is the change in the electric potential energy of the charge? (a) -2.4 × 10-3 J (d) +1.5 ×10-4 J (b) -1.5 ×10-4 J (e) +2.4 ×10-3 J (c) zero joules X

14. P and Q are points within a uniform electric field that are separated by a distance of 0.1 m as shown. The potential difference between P and Q is 50 V. 1. Determine the magnitude of this electric field. (a) 0.5 V/m (c) 50 V/m (e) 5000 V/m (b) 5.0 V/m (d) 500 V/m 2. How much work is required to move a +1000 mC point charge from P to Q? (a) 0.02 J (c) 200 J (e) 5000 J (b) 0.05 J(d) 1000 J X X

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

16. Electric Potential of Multiple Point Charges • Superposition principle applies • The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges • The algebraic sum is used because potentials are scalar quantities

17. Electrical Potential Energy of Two Charges • V1 is the electric potential due to q1 at some point P1 • The work required to bring q2 from infinity to P1 is q2V1 • This work is equal to the potential energy of the two particle system P1

18. Notes About Electric Potential Energy of Two Charges • If the charges have the same sign, PE is positive • Positive work must be done to force the two charges near one another • The like charges would repel • If the charges have opposite signs, PE is negative • The force would be attractive • Work must be done to hold back the unlike charges from accelerating as they are brought close together

19. Q Q E=0 Question Is there a point along the line joining two equal positive charges where the electric field is zero? Where the electric potential is zero? Answer: The sum of both potentials is nonzero at any finite distance but becomes zeroat infinite distance.

20. 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

21. Summary • Electric potential: PE per unit charge, V=DPE/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=∞)