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

• Wexternal agent increases the potential energy

• Wexternal agent = -Wfield

• U = - Wfield

• Wfield = ∫F·ds

Ch 25: Electric Potential25.1 Potential Difference and Electric

Potential

ds

q0E

• Electric Potential V = U/q0

• Volts = Joules/Coulomb

Ch 25: Electric Potential25.1 Potential Difference and Electric

Potential

P25.1 (p. 714)

Ch 25: Electric Potential25.2 Potential Differences in a Uniform

Electric Field

A

B

dds

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 Potential25.3 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 Potential25.4 Obtaining the Value of E from V

Ex = -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

AB

C

D

CT7:The electric potential for this quadrupole is non-zero at point

A. A

B. B

C. C

D. D

AB

C

D

Ch 25: Electric Potential25.5 Electric Potential Due to Continuous

Charge Distributions

P25.37 (p. 717)

Revisit P23.27 (p. 668)

qi dq Vi ∫dV

dl = rd

dq = dl

r

dV

Ch 25: Electric Potential25.6 Electric Potential Due to Charged

Conductor

Fig 25-22, p.778

Example 25.9

1/2 = r2/r1

E1/E2 = r2/r1

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 conductoran equipotential surface?

V = ∫E·ds = 0 because E ds.

E = 0 in a cavity in a conductor

SummaryApproach

Type of

theory

Force

(vector)

Energy

(scalar / only differences matter)

Action-at-a-distance

FE UE

Field/Potential E = FE / q0 VE = UE / q0

Ex = -V / x etc.V = -Eds