Fig 25-CO, p.762

Concept Question 1: At which point is the positive charge at a lower potential energy? A. Point A. B. Point B. C. Both A and B are at the same potential energy. +

Energy Review W external agent increases the potential energy W external agent = -W field U = - W field W field = ∫ F·ds Ch 25: Electric Potential 25.1 Potential Difference and Electric Potential ds q0Eq0E

Electric Potential V = U/q 0 Volts = Joules/Coulomb Ch 25: Electric Potential 25.1 Potential Difference and Electric Potential P25.1 (p. 714)

Ch 25: Electric Potential 25.2 Potential Differences in a Uniform Electric Field A B d ds

Concept Question 3: At which point is the charge at a lower potential energy? A. Point A. B. Point B. C. Both A and B are at the same potential energy. D. It depends on the sign of the charge.

Concept Question 2: Which point is at a lower potential? A. Point A. B. Point B. C. Both A and B are at the same potential. D. It depends on the sign of the charge.

P25.8 (p. 715)

Concept Question 4: The integral of the scalar product of E and ds along the path A. depends on the path. B. is independent of the path. C. depends only on B. D. depends only on A. E. doesn’t depend on either A or B.

dr Ch 25: Electric Potential 25.3 Electric Potential and Potential Energy Due to Point Charges +q E B A The radial direction is outward

Discuss BCA: Give a physical explanation of the fact that the potential energy of a pair of charges with the same sign is positive, whereas the potential energy of a pair of charges with opposite signs is negative.

P25.15 (p. 716) P25.19 (p. 716)

Ch 25: Electric Potential 25.4 Obtaining the Value of E from V E x = -dV/dx, etc. P25.30 (p. 717)

Why are electric equipotential lines perpendicular to electric field lines? Because then  V = ∫ E·ds = 0.

Hint: Think of the definition of work and potential. A. B. C. Concept Question 5

CT6: The E field for this quadrupole is stronger at point A. A B. B C. C D. D A B C D

CT7: The electric potential for this quadrupole is non-zero at point A. A B. B C. C D. D A B C D

Ch 25: Electric Potential 25.5 Electric Potential Due to Continuous Charge Distributions P25.37 (p. 717) Revisit P23.27 (p. 668) q i  dq V i  ∫ dV

dl = rd  dq = dl  r dV

Ch 25: Electric Potential 25.6 Electric Potential Due to Charged Conductor

Fig 25-22, p.778

Example 25.9  1 / 2 = r 2 /r 1 E 1 /E 2 = r 2 /r 1 recall E = / 0 just outside a conductor  and E are bigger on surfaces of smaller radius of curvature

Fig 25-21, p.778 Why is the surface of a conductor an equipotential surface?  V = ∫ E·ds = 0 because E  ds.

E = 0 in a cavity in a conductor

Summary Approach Type of theory Force (vector) Energy (scalar / only differences matter) Action-at-a- distance FEFE UEUE Field/PotentialE = F E / q 0  V E =  U E / q 0 E x = -V / x etc. V = -  E  ds