Electric potential energy:

.( : : ) :

.f i

Electrostatic force does work on a particleW F s qE sPotential energy i initial state f final state

U U U W

•

= ⋅ = ⋅•Δ = − = −

v vv v

Electric potential : : arg .

( ) : ( ) , .( )

UV potential energy per unit ch eq

J Joule N N m J VUnit V volt EC Coul C C m C m m

Δ• Δ ≡

⋅= = = = =

⋅ ⋅

E

f

i

q

191 1 1.6 10eV e V J−= ⋅ = ×

Work done by applied force .appW -W• =

sourceq testqE♁ ♁

000app

WU

W

>Δ <

<

000app

WU

W

<Δ >

>

♁

sourceq testqE

000app

WU

W

<Δ >

>

000app

WU

W

>Δ <

<

In chapter 8 we defined theassociated with a conservative force as the negative value of the work that the force must do on a particle to take it from an initial posi

U

W

Δchange in potential energy

tion to a final position .

( )f

i

i f

x

f ix

x x

U U U W F x dxΔ = − = − = − ∫

O

x. . .xi x xf

F(x)

( )f

i

x

x

U F x dxΔ = − ∫

(24 - 2)

Electric Potential Energy :

A

B

Consider an electric charge moving from an initial position at point A to a final position at point B under the influence of a

known electric field . The force exerted on the charge is:

o

o

q

E F q E=r r r

f f

oi i

U F ds q E dsΔ = − ⋅ = − ⋅∫ ∫r rr r f

oi

U q E dsΔ = − ⋅∫r r

(24 - 2)

O

Definition of voltage :

Units of : J/C known as the "Volt"

Consider a pont charge placed at the origin.We will use the defini

o

Wq

q

V

V

Δ = −

SI Units of

Potential due to a point charge

V :

P

tion given in the previous page to determine the potential at point P a distnce fro m O.

VR

14P

o

qVRπε

=

(24 - 4)

O

2

22

cos0

The electric field generated by is: 4

4

1 14

1

4

R

PR R

o

Po R

PRo o

V E ds Edr Edr

qq Er

q d drVr

q qVr R

rx x

πε

πε

πε πε

∞ ∞

∞

∞

∞

= − ⋅ = =

=

=

⎡ ⎤→ = − =⎢ ⎥⎣ ⎦

= −

∫ ∫ ∫

∫∫

r r

14P

o

qVRπε

=

(24 - 4)

V=V0

V=V0

Ev

Ev

tan

tan tan

Equi-potential surface

0

( , 0.

00 0)

Calculating V from E:

f f

i i

U WVq q

E Equipotential surfaceIf not E

q E FF S W V

dW F ds qE ds

W dW qE ds q E ds

V

•Δ

Δ = = − =

⇒ ⊥∃ ≠

⇒ ⋅ = ≠

⇒ ⋅ = ≠ ⇒ Δ ≠

•

= ⋅ = ⋅

= = ⋅ = ⋅

Δ = −

∫ ∫ ∫

v

v vv w

v vv v

cosf f

f ii i

W E ds V V E dsq

θ= − ⋅ = − = −∫ ∫v r

1 21

21

2

1 2

0. (V , 0)

V V V

E ds

E ds

EdV W

Δ = −

= − ⋅

= −

= − << >

∫∫

v vEX.

2

1

+

_d

q⊕E ds

2

1

+

_

dq⊕

E ds

2 12

12

12 2

1 1

2 1

cos180

0

( , 0)

V V V

E ds

E ds

Eds E ds Ed

V V W

Δ = −

= − ⋅

= − °

= = = >

> <

∫∫∫ ∫

v v

e-

+V~0

+V~20kV

v=?

2

19 2

7

12

1( 1.6 10 ) 20218.4 10 /4

e

W U K q V mv

k m v

v m s c

−

= −Δ = Δ ⇒ − ⋅Δ =

− − × ⋅ =

⇒ = × ≈

EX.

Potential of group of point charges•

2R 3R1R

2q 3q1q

?V =

0 i

i i

k qVR⋅

=∑

Potential of continuous charge distribution•

rQΔ

Q

Ο

?=V0

0

0 0

0

0 limQ Q

k QVrk QV

rk Q k dQV

r rΔ→

ΔΔ =

Δ=

ΔΔ→ ⇒ = =

∑

∑ ∫

r1r2

r3

P

q1

q3

q2

Consider the group of three point charges shown in the figure. The potential generated by this group at any point P is calculated using the principle of superposition

We determine th

V

1. 1 2 3

1 2 31 2 3

1 2 3

e potentials , and generated by each charge at point P.

1 1 1 , , 4 4 4o o o

V , V V

q q qV V Vr r rπε πε πε

= = =

(24 - 5)1 2 3V V V V= + +

Potential due to a group of point charges

r1r2

r3

P

q1

q3

q2

1 2 3

1 2 3

1 2

11 2

1 2 3

We add the three terms:

The previous equation can be generalized for charge

1 1 1 4 4 4

1 1 1 1

s as follo

...4 4 4

w

4

s:

o o o

nn i

o o o n o i

q qV V V V

n

qVr r r

q q q qVr r r r

πε πε πε

πε πε πε πε

= + +

= + + + =

= + +

∑

2.

(24 - 5)1 2 3V V V V= + +

EX. Potential for uniformly charged thin ring

Qr z

R

φddQ

?)( =ZV 0

20

0

200

0

0

2 2

, 2 2

2

2

Q

k dQ Q QV dQ Rd dr R

k Q dr

k Q dr

k Qr

k QR z

π

π

ϕ ϕπ π

ϕπ

ϕπ

= = ⋅ =∫

= ⋅∫

= ∫

=

=+

A

B

C

pr Consider the electric dipole shown in the figure.We will determine the electric potential created at point P by the two charges of the dipole using superposition.Point P is at a distance from the

V

r

( ) ( )( ) ( )

( ) ( ) ( ) ( )

center O of the dipole.Line OP makes an angle with the dipole axis

14 4o o

r rq q qV V Vr r r r

θ

πε πε− +

+ −+ − − +

⎛ ⎞ −= + = − =⎜ ⎟⎜ ⎟

⎝ ⎠

(24 - 6)

Example :

Potential due to an electric dipole

A

B

C

pr( ) ( )

2( ) ( ) 2 2

We assume that where is the charge separation. From triangle ABC we have: cos

cos 1 cosAlso: 4 4

where the electric dipole moment.o o

r d d

r r d

q d pr r r Vr r

p qd

θ

θ θπε πε

− +

− +

− ≈

≈ → ≈ =

= =

?

(24 - 6)

2

1 cos4 o

pVr

θπε

=

:V from E gDeterminin•

sdρ

⊕q

Eρ

dVsdEqsdFdW ρρρρ

⋅=⋅=

dVsdEq

dW−=⋅=

ρρ

) Calculus. of principle lFundamenta

(

path toparallel field ,||

||

||

∫−=−⇔

−=

dsEVV

dsdVEdsEif

if

kEjEiEE zyxˆˆˆ

general,In

++=ρ

,xVEx ∂∂

−= ,yVEy ∂∂

−= ,zVEz ∂∂

−= .)ˆˆˆ()ˆˆˆ( Vz

ky

jx

ikzVj

yVi

xVE

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

−=ρ

,ˆˆˆ)(

kz

jy

ixdel ∂

∂+

∂∂

+∂∂

≡∇ .VE −∇=ρ

EX. Check field for single charge

rqkV 0=

r

q

20

|| rqk

drdV

dsdVE =−=−=

rr

qkE ˆ

20=

ρ

EX. Check filed for uniformly charged thin ringknown

Q

V 0=∂∂

−=xVEx 0=

∂∂

−=yVEy

....=−=∂∂

−=dzdV

zVEz

dqO A

2 2

Potential created by a line of charge oflength L and uniform linear charge densityλ at point P. Consider the charge element

at point A, a distance from O.From triangle OAP we have:

dq dx x

r d x

λ=

= + Here is the distance OPThe potential created by at P is:

ddV dq

(24 - 8)

Example :

dqO A

(24 - 8)

( )( )( )

2

2 2

2 20

2 2

0

2

2

2

2

2

1 14 4

4

ln

l

4

n ln4

n

l

o o

L

o

L

o

o

dx x d

dq dxdVr d x

dxVd x

V x d x

V x

d

L

x

L

x

d

λπε πε

λπε

λπελπε

= =+

=+

⎡ ⎤= + +⎢ ⎥⎣ ⎦

⎡ ⎤= +

= + +

+ −⎢ ⎥⎣ ⎦

+∫

∫

Lets assume that the electric field forms an angle with thepath . The work done by the electric field is:

cos cosWe also know that 0. Thus: cos 0

0, 0, 0

Er

W F r F r qE rW qE r

q E r

θ

θ θθ

Δ

= ⋅Δ = Δ = Δ= Δ =

≠ ≠ Δ ≠

r

r

r r

Thus: cos 0 The correct picture is shown in the figure belo

9w

0θθ = → = °

(24 - 11)

SV

Er

Examples of equipotential surfaces and the corresponding electric field lines

Uniform electric field Isolated point charge Electric dipole

(24 - 12)

A

B

V V+dV

Now we will tackle the reverse problem

i.e. determine if we know . Consider two equipotential surfaces thatcorrspond to the values and separated by a distance as shown in the figure.

E V

V V dVds

+

r

Consider an arbitrary direction represented by the vector . We will allow the electric field to movea charge from the equipotenbtial surface to the surface .

o

ds

qV V dV+

r

(24 - 13)

E VCalculating the electric field from the potential r

The work done by the electric field is given by: (eqs.1)

also cos cos (eqs.2)If we compare these two equations we have:

cos cos

From triangle PAB we see that

o

o

o o

W q dVW Fds Eq ds

dVEq ds q dV Eds

θ θ

θ θ

= −= =

= − → = −

cos is the component of along the direction s.

Thus:

s

s

EE E

VEs

θ

∂= −

∂

r

sVEs

∂= −

∂(24 - 13)

A

B

V V+dV