§8-5 Electric potential Electric potential

Electrostatic field does work for moving charge--E-field possesses energy

1.Work done by electrostatic force Let test charge q0 moves a b along arbitrary path in the E-field set up by point charge q .

The work done by electrostatic force=?

c

a

b

q

ar

br

q0

ldFdW

ldEq

0

dlEq cos0

Edrq0

ld

: displacement

ld

c

a

b

q

ar

br

dr

r E

b

aoab EdrqW

b

a

r

rdr

r

qq2

0

0 1

4

)11

(4 0

0

ba rr

qq

The work depends on only the initial position and final position of qo, and has nothing to do with the path.

c

a

b

q

ar

br

q0

b

aab ldEqW

0

ibia

n

i

i

rr

qq 11

41 0

0

b

a n ldEEEq

)( 210

---has nothing to do with path

When qo moves in the E-field set up by charges’ system q1, q2, qn ,

n

iiW

1

When qo moves in the E-field set up by charged body,

Conclusion:-- the work has nothing to do with path

2. Circular theorem of electrostatic field

When q0 moves along a closed path L ，

00 LLldEqdWW

Electrostatic force is conservative force.

E-force does work:

3. Electric potential energy

b

a

ab ldEqW

0

-- The E-potential energy when q0 at point a and b.

PbPa EE

PbEPaE 、

Electrostatic field is conservative field.

The work done by electrostatic force = the decrease of the electric potential energy

0 ldEL

Circular theorem of electrostatic field

q0 moves in E-field a b ，

Notes

(1) EP a is relative quantity. If we want to decide the

magnitude of EP when q0 at a point , we must choose zero reference point of E-potential energy.

PaE a

ldEq

0

Z

PEW

The work done by E-force for q0=- increment of E-potential energy.

Z--zero E-potential energy pointZ 0PE

The choice of zero E-potential energy point ： Choose zero point at when the charge distr

ibution is finite.

Choose zero point at the finite distance point w

hen the charge distribution is infinite.

(2) EP is scalar. It can be positive, negative or zero.

0q(3) EP depends on E-field and ， it belongs the system.

0q

EU Pa

a

4. Electric potential

a

ldEZ

Definition

b

a

ldE

E-potential difference:

0q

EEUUV PbPa

baab

)(0 baab UUqW

--Describe the character of E-field.

Work:

5.Calculation of E-potential(1) The E-potential of a point charge q

aa ldEU

r

drr

q2

04

1

a

Edr

r

q

04

1

Take U=0 ， then the E-potential of a point a :

The distance from q to a

DiscussionU

r+

U

r

If q>0 ， U>0 for any point in the space.

when r ， U U() 0

If q<0 ， U<0 for any point in the space.

when r ， U U() 0

r

1

aa ldEU

a naa

ldEldEldE

21

nUUU 21

n

iiU

1

n

i i

i

r

q

1 04

（ 2 ） The E-potential of a system

The system of point charges ： q1,q2,,qn

qi

a

ri

q1

r1

--superposition principle of E-potential

dq

ar

（ 3 ） The E-potential of charged body

Divide q many of dq

For any dq ：

r

dqdU

04

1

For entire charged body ：

charge element

r

dqdUUa

04

1

Integrating for charged body

Caution ！！

This method can be used for the finite distribution charged body.

[Example 1] Four point charges q1 ＝ q2 ＝ q3 ＝ q4

＝ q is put on the vertexes of a square with edge of a respectively.

Calculate (1)The E-potential at point 0. (2)If test charge q0 is moved from to 0, how much work does the E-force do?

2

ar

1q 2q

3q4q

0a

a

a

a

6. Examples of calculating E-potential

a

qa

q

r

qUU

ii

00

0

4

10

2

2

1

4

14

（ 1 ）

a

qqUqEEW PP

0

0000

2

（ 2 ）

Definition method

a

a ldEUZ

Integrating for path

Calculate the E-potential set up by a charged body

Two methods

Use the definition of E-potential as the distribution of is known. – definition methodE

Use the superposition principle of E-potential-- superposition method.

Integrating for charged body

r

dqU

04q

superposition method

xPxrR

0

[Example 2] Calculate the E-potential on the axis of a uniform charged ring. q 、 R are known.Solution

Method Use the E-field distribution of the ring that was calculated before.

23

220 )(4 Rx

qxE

The direction: along x axis

P

ldEU

xdx

xR

qx2/32

0 )(4

x

dxE

2204 xR

q

r

dqdU

04

1

Method -- superposition method

q dq

dUU

2204

1

xR

q

qdq

r04

1

xPxrR

0

U

x0

Discussion

(1) ox

R

qU

00 4

1

Rx(2) x

qU

04

1

If the charged body is a half circle, ？0U0

R

R

+

++++

+

++

q

[Example 3] Calculate the E-potential distribution of a charged spherical surface

E-field distribution:

E0

204

1

r

q

Rr（ ）

Rr（ ）

Solution

Definition method

Zero potential point ：

R

+

++++

+

++

q

P

ldEU

drr

q

r2

04

1

(1) For any point P outside the sphere

r

q

04

1

Rr（ ）

Pr

(2) For any point inside the sphere surface

+

++++ q

++

+

+

++

+

R

P

ldEU

R

r

dr0

P

rdr

q

R2

04

1

R

q

04

1

The sphere is equipotential.

Rr

分段积分

Conclusion

R

q

04

1

Rr（ ）

r

q

04

1

Rr（ ）

U

The distribution curve of E-potential

The distribution curve of E-field ： E

R

R r

rO

O

88

r2

r

1

1

superposition method ：

Integrating for charged body

r

dqU

04q

It’s very complex ！！

Conclusion

When the E-field distribution is symmetry and it can be calculated by using Gauss’s Law conveniently, it is simpler to calculate potential by using definition method.

When the E-field distribution is not symmetry an

d it can’t be calculated by using Gauss’s Law conveniently, it is simpler to calculate potential

by using superposition method.

P

[Example 4] Calculate the E-potential distribution of an infinite line with uniform charge(the linear density isλ).

Solution

Use definition method

rE

02

pr

How to choose the zero potential point?

Choose any point b as zero potential point

b

PP ldEU

When rp<rb ， U >0. When rp>rb ， U <0.

b

P

P

b

r

rln

2 0

Finite distance to the charged line

§8-6 Equipotential surface and Potential gradient

1. Equipotential surface

--the potential has the same value at all points on the surface.

Positive point chargeElectric dipole

Dash line-- equipotential surface

Real line-- -lineE

A parallel plate capacitor

++ ++++ +++

0)(0 baab UUqW

Prove ： Assume q0 moves along equipotential surface a b ，then ：

The properties of equipotential surfaces

No net work is done by the E-field as a charge moves between any two points on the same equipotential surface.

ab

q0

ldEqdW

0

Prove ：

0

ldE

EAssume on an equipotential surface, the field

at point P is

then ：

q0 moves along equipotential surface, ld

q0Pld

E

-lines are always normal to equipotential surfacesE

b

aba ldEUU

E

a b

Prove ：Assume there are two equipotential surface U a , U b

then

ld

b

adlE 0

ba UU

-line points on the direction of the increase of the potential.

E

Ua Ub

E

Uc

r2

r1

the density of equipotential surfaces shows the magnitude of E-field.Prove ：

Assume there is a family of equipotential surfacesUa 、 U b 、 Uc 、 E1

E2

then ：11 rEUU ba

22 rEUU cb

1

2

2

1

r

r

E

E

2. Potential gradient

Equipotential surface

P

P ldEU

-lineE

If the distribution of U is known, How can we calculate ? E

(1) Special example ： uniform field

unit positive charge ab, the work done by E-force:

ba UUW nE

nEU

n

UE

E

C

l

na b

nE

Express that the ’s direction is the direction of U decrease.

E

q=1

q=1 moves ac , the work done by E-force ：

lE cosca UUW

lEU l

l

UEl

lEl

’s component at the direction of l

E

’s component at any direction = the negative magnitude of the rate of U change with distance on that direction.

E

(2) Any E-field:

U

dUU 0dU

nd

ld

dl

dUEl

dn

dUE

definition ：potential gradient

grad U ndn

dU

ndn

dUE

E

grad U

n

The normal direction of equipotential surface,point on U increasing

= U

U

In Cartesian coordinate system

x

UE x

y

UE y

z

UEz

E

)( kz

Uj

y

Ui

x

U

a.the magnitude of at any point = the maximum of the rate of U change with distance on that point. the direction of is perpendicular to the equipotential surface at that point and along the the direction of U decreasing.

E

E

b. In the space that U is constant, U = 0 ， E = 0

c. E is not sure = 0 in the space of U= 0.

U is not sure = 0 in the space of E= 0.

conclusion

[Example 1] Calculate the E-field of an electric dipole.

l

rr

r

x

y

q q

P( )x y

r

q

r

qU

o

o

4

1

4

1

2rrr coslrr

24

cos

r

pU

o

21

22 yx

x

r

xcos

23

220 )(4 yx

pxU

25

220

22

)(4

)2(

yx

yxp

x

UE x

25

220 )(4

3

yx

pxy

y

UE y

E

jEiE yx

R

P xx0

[example 2] calculate the E-field on the axis of the

round charged plate using its potential gradient .

Solution The potential of any ring on the axis:

r

dqdU

04

1

dsdq d2

22 xr 2202 x

dd

R

P xx0

dr

Ed

xxRU 22

02

i.e., the potential on the axis: xUU

idx

dUE

220

12 xR

x

R

x

ddUU

0 2202

§8-7 E-Force Exerted on a Charge§8-7 E-Force Exerted on a Charge

1. The E-force exerted on a charged particle

The torque exerted on the dipole:

EqF

EPM e

sinlFM

F

F

q

+q

l

sin l qE sinEPe

lqPe

--electric dipole moment

(1) E-dipole in the uniform field

The E-dipole will rotate under the action of the torque until the direction of is the same with E

eP

F

F

q

+q

l

Assume the potential energy of E-dipole is zero at the position =900 Ep (=900) =0

Then the potential energy of E-dipole at any orientation is

00

90)90(sin dEPWE ep

cosEPe

EPE ep

(2) E-dipole in non-uniform field

21 FFF

2211 sin2

sin2

ee rF

rFM

)( 21 EEq

ee

ee r

EEP

r

EEqr 2121

F1

F2

q

+q

1

l

2

0 --Move to the area of larger E-field

0 --rotating

2. A moving charged particle in a uniform E-field

EqF

m

qEa x

m

qExavv 222

02

qUmvmv 20

2

2

1

2

1

Ev

//)1( E

vq

td

vdmam

Ex=U

22

2

1

2

1t

m

qEatx

20

2

2

1

v

yE

m

qx

tvy 0

Ev 0)2(

E

0v

q x

yv