Chapter 23

Electric Potential

Physics for Scientists & Engineers, 3rd EditionDouglas C. Giancoli

© Prentice Hall

10P04 -

Potential Energyand Potential

Start with Gravity

11P04 -

Gravity: Force and Work

Gravitational Force on m due to M:

2ˆg

MmGr

= −F r

Work done by gravity moving m from A to B:

g

B

g AW d= ⋅∫ F s PATH

INTEGRAL

12P04 -

Work Done by Earth’s GravityWork done by gravity moving m from A to B:

ggW d= ⋅∫F s

( )2ˆ ˆˆ

B

A

GMm

rdr rdθ−⎛ ⎞= ⋅ +⎜ ⎟

⎝ ⎠∫ r r θ

1 1

B A

GMmr r

⎛ ⎞= −⎜ ⎟

⎝ ⎠

2

B

A

r

r

GMm drr

= −∫B

A

r

r

GMm

r= ⎡ ⎤⎢ ⎥⎣ ⎦

What is the sign moving from rA to rB?

13P04 -

Work Near Earth’s Surface

2 ˆ ˆE

GM gr

≈ − = −g y yG roughly constant:

Work done by gravity moving m from A to B:

( )ˆB

Amg d= − ⋅∫ y sg gW d= ⋅∫F s

By

yAmgdy= −∫ ( )B Amg y y= − −

Wg depends only on endpoints– not on path taken –Conservative Force

14P04 -

Potential Energy (Joules)B

g B A g g extAU U U d W W∆ = − = − ⋅ = − = +∫ F s

02ˆ(1) g g

GMm GMmU Ur r

= − → = − +F r

• U0: constant depending on reference point• Only potential difference ∆U has

physical significance

0ˆ(2) g gmg U mgy U= − → = +F y

15P04 -

Gravitational Potential(Joules/kilogram)

Define gravitational potential difference:

g( / )B Bg

g A A

UV m d d

m∆

∆ = = − ⋅ = − ⋅∫ ∫F s g s

gFieldForce Energy Potential

Just as , g gU V→ ∆ → ∆F g

That is, two particle interaction single particle effect

17P04 -

Move to Electrostatics

19P04 -

Potential & Energy

B

A

V d∆ ≡ − ⋅∫E sUnits:

Joules/Coulomb = Volts

Work done to move q from A to B:

ext B AW U U U= ∆ = −

q V= ∆ Joules

18P04 -

Gravity - Electrostatics

2ˆMG

r= −g r

Mass M Charge q (±)

2ˆe

qkr

=E r

g m=F g E q=F EBoth forces are conservative, so…

B

g gAU d∆ = − ⋅∫ F s

B

EAU d∆ = − ⋅∫ F s

B

AV d∆ = − ⋅∫ E s

B

g AV d∆ = − ⋅∫ g s

20P04 -

Potential: Summary Thus Far

Charges CREATE Potential Landscapes

0 0"0"

( )V V V V d= + ∆ ≡ − ⋅∫r

r E s

21P04 -

Potential Landscape

Positive Charge

Negative Charge

22P04 -

Potential: Summary Thus Far

Charges CREATE Potential Landscapes

0 0"0"

( )V V V V d= + ∆ ≡ − ⋅∫r

r E s

Charges FEEL Potential Landscapes

( ) ( )U qV=r r

We work with ∆U (∆V) because only changes matter

Figure 23-4

Figure 23-5

26P04 -

Potential Created by Pt Charge

θrsd ˆˆ θdrdr +=

B

B A AV V V d∆ = − = − ⋅∫ E s

2

ˆr

kQ rE =

2 2

ˆB B

A A

drkQ d kQr r

= − ⋅ = −∫ ∫r s

1 1

B A

kQr r

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Take V = 0 at r = ∞:

rkQrV =)(ChargePoint

28P04 -

Potential Landscape

Positive Charge

Negative Charge

Figure 23-8

Figure 23-7

Figure 23-9

42P05 -

Potential from E

45P05 -

Potential for Uniformly Charged Non-Conducting Solid Sphere

43P05 -

Potential for Uniformly Charged Non-Conducting Solid SphereFrom Gauss’s Law

20

30

ˆ,4

ˆ,4

Q r Rr

Qr r RR

πε

πε

⎧ >⎪⎪= ⎨⎪ <⎪⎩

rE

r

B

B AA

V V d− = − ⋅∫E sUse

( )0

BV V=

− ∞Region 1: r > a

204

r Q drrπε∞

= −∫0

14

Qrπε

=

Point Charge!

44P05 -

Potential for Uniformly Charged Non-Conducting Solid Sphere

( ) ( )R r

RdrE r R drE r R

∞= − > − <∫ ∫

Region 2: r < a( )

0

DV V=

− ∞

2

20

1 38

Q rR Rπε⎛ ⎞

= −⎜ ⎟⎝ ⎠

2 30 04 4

R r

R

Q Qrdr drr Rπε πε∞

= − −∫ ∫

( )2 23

0 0

1 1 14 4 2

Q Q r RR Rπε πε

= − −

Figure 23-6 (a)

Figure 23-6 (b)

46P05 -

Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). If V=0 at x=0 (definition) then what is V(x) for x>0?

x̂

47P05 -

Group Problem: Spherical ShellsThese two spherical shells have equal but opposite charge.

Find E everywhere

Find V everywhere (assume V(∞) = 0)

Figure 23-11

Figure 23-12

Figure 23-36

Figure 23-27

Figure 23-23

30P04 -

Deriving E from V

ˆx∆ = ∆s i

A = (x,y,z), B=(x+∆x,y,z)

B

A

V d∆ = − ⋅∫E s

( , , )

( , , )

x x y z

x y z

V d+∆

∆ = − ⋅∫ E s ≅ − ⋅∆E s ˆ( ) xx E x= − ⋅ ∆ = − ∆E i

xV VEx x

∆ ∂≅ − → −

∆ ∂Ex = Rate of change in V

with y and z held constant

31P04 -

Deriving E from VIf we do all coordinates:

V= −∇E

ˆ ˆ ˆV V Vx y z

⎛ ⎞∂ ∂ ∂= − +⎜ ⎟∂ ∂ ∂⎝ ⎠

E i + j k

ˆ ˆ ˆ Vx y z

⎛ ⎞∂ ∂ ∂= − +⎜ ⎟∂ ∂ ∂⎝ ⎠

i + j k

Gradient (del) operator:

ˆ ˆ ˆx y z∂ ∂ ∂

∇ ≡ +∂ ∂ ∂i j+ k

34P04 -

Configuration EnergyHow much energy to put two charges as pictured?

1) First charge is free2) Second charge sees first:

1 2

1212 2 2 1

14 o

q qU W q V

rπε= = =

35P04 -

Configuration EnergyHow much energy to put three charges as pictured?

1) Know how to do first two2) Bring in third:

( )3 3 1 2W q V V= + 3 1 2

0 13 234q q q

r rπε⎛ ⎞

= +⎜ ⎟⎝ ⎠

Total configuration energy:

1 3 2 31 22 3 12 13 23

0 12 13 23

14

q q q qq qU W W U U Ur r rπε

⎛ ⎞= + = + + = + +⎜ ⎟

⎝ ⎠

36P04 -

In Class Problem

How much energy in joules is required to put the three charges in the configuration pictured if they start out at infinity?

What is the electric potential in volts at point P?

Suppose you move a fourth change +3Q from infinity in to point P. How much energy does that require (joules)?