The divergence of EIf the charge fills a volume, with charge per unit volume .

'dq d

Where d is an element of volume.

For a volume charge:

''

20

1 ( )ˆ( )

4 v

rE R rd

r

R

Thus:

3 ' ' '

0

1. 4 ( ) ( )

4 v

E r r r d

0

1. ( )E r

Gauss’s law in

differential form.

32

ˆ. 4 ( )

rr

r

' '2

0

ˆ1. . ( )

4 v

rE r d

r

Spherical polar coordinates (r, , )

r: the distance from the origin: the angle down from the z axis is called polar angle: angle around from the x axis is called the azimuthal angle

sin cosx r sin siny r

cosz r

ˆ ˆ ˆ ˆsin cos sin sin cosr x y z ˆ ˆ ˆ ˆcos cos cos sin sinx y z

ˆ ˆ ˆsin cosx y ˆ ˆˆ sindl drr rd r d

The Curl of E

For a point charge situated at origin:

20

1ˆ

4

qE r

r

Line integral of the field from some point a to some other point b:

In spherical polar coordinates,

ˆ ˆˆ sindl drr rd r d

20

1.

4

qE dl dr

r

20

1.

4

b b

a a

qE dl dr

r

0

1

4 a b

q q

r r

True for electrostatic field.

. 0E dl Apply stokes theorem:

0E

The integral around a closed path:

Electric PotentialBasic concept:

The absence of closed lines is the property of vector field whose curl is zero.

E is such a vector whose curl is zero.

Using this special kind of it’s property we can reduce a vector problem: using V, we can get E very easily.

Vector whose curl is zero, is equal to the gradient of some scalar function

E=0 the line integral of E around any closed loop is zero (due to Stokes' theorem).

E V

otherwise you could go out along path (i) and return along path (ii) and

Because the line integral is independent of path, we can define a function

O is some standard reference point.

. 0E dl Therefore the line integral of E from point a to point b is the same for all paths.

. 0E dl

( ) .r

o

V r E dl is called electric potential

The potential difference between two points a and b:

( ) ( ) . .b a

o o

V b V a E dl E dl

. .b o

o a

E dl E dl

.b

a

E dl Using fundamental theorem for gradients:

( ) ( ) ( ).b

a

V b V a V dl So ( ). .

b b

a a

V dl E dl E V

Electric field is the gradient of a scalar potential.

Electric Potential at an arbitrary point

•Electric potential at a point is given as the work done in moving the unit test charge (q0) from infinity (where potential is taken as zero) to that point.

• Electric potential at any point P is

Note that Vp represents the potential difference dV between the point P and a point at infinity.

S.I. unit J/C defined as a volt (V) and 1 V/m = 1 N/C

.p

pV E ds

0

p

WV

q

Potential Difference in Uniform E field

• Electric field lines always point in the direction of decreasing electric potential.

Example: Uniform field along –y axis (E parallel to dl)

.B B

B A

A A

V V V E dl Edl B

A

V Edl Ed • When the electric field E is directed downward,

point B is at a lower electric potential than point A. A positive test charge that moves from point A to point B loses electric potential energy.

Potential Diff. in Uniform E field

Charged particle moves from A to B in uniform E field.

q

.b

a

V E ds

Potential Diff. In Uniform E field (Path independence)

Show that the potential difference between point A and B by moving through path (1) and (2) are the same as expected for a conservative force field.

By path (1), . cosB

A

V E dl El

path (2)

= 0 since E is to dl

. cosC

A

V E dl Eh El

. .C B

A C

V E dl E dl

Equipotential Surfaces (Contours)

VC = VB ( same potential)

In fact, points along this line has the same potential. We have an equipotential line.

. 0B

C

V E dl

No work is done in moving a test charge between any two points on an equipotential surface.

The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field.

Equipotential Surface

Equipotential Surfaces (dashed blue lines) and electric field lines (orange lines) for (a) a uniform electric field produced by infinite sheet of charge, (b) a point charge, and (c) an electric dipole. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.

16

Electrostatic Potential of a Point Charge at the Origin

Q

P

r

'2'

0

2'0 0

4

4 4

r r

r

QV r E dl dr

r

Q dr Q

rr

17

Electrostatic Potential Resulting from Multiple Point Charges

Q1

P(R, ,q f)

r 1R

1rO

Q2

2r

1 04

nk

k k

QV r

R

2R

18

Electrostatic Potential Resulting from Continuous Charge Distributions

0

0

0

1

4

1

4

1

4

L

S

V

dlV r

R

dsV r

R

dvV r

R

line charge

surface charge

volume charge

19

Charge Dipole

• An electric charge dipole consists of a pair of equal and opposite point charges separated by a small distance (i.e., much smaller than the distance at which we observe the resulting field).

d

+Q -Q

Dipole Moment

• Dipole moment p is a measure of the strength of the dipole and has its direction.

p Qd+Q

-Q

dp is in the direction from the negative point charge to the positive point charge

21

Electrostatic Potential Due to Charge Dipole

observationpoint

d/2

+Q

-Q

z

d/2

q

P

ˆzp a Qd

R

Rr

22

d/2

d/2

q

cos)2/(

cos)2/(

22

22

rddrR

rddrR

R

r

P

0 0

,4 4

Q QV r V r

R R

R

• first order approximation from geometry:

cos2

cos2d

rR

drR

d/2

d/2

q

lines approximatelyparallel

R

R

r

24

• Taylor series approximation:

cos2

111

cos2

11

cos2

11

cos2

111

r

d

rR

r

d

r

r

d

r

dr

R

1,11

:Recall

xnxx n

20

0

4

cos

2

cos1

2

cos1

4,

r

Qd

r

d

r

d

r

QrV

25

• In terms of the dipole moment:

20

ˆ

4

1

r

apV r

Electric Potential Energy of a System of Point Charges

1

0

1

4

qV

r2 ( )W q V r

1 3 2 31 212 13 23

0 12 13 23

1( )

4

q q q qq qW W W W

r r r

2W F r q E r

q1

q22

b b

a a

W F dl q E dl

2[ ( ) ( )]W q V b V a

2[ ( ) ( )]W q V r V

and we know

The Energy of a Continuous Charge Distribution

For a volume charge density p, 1

2W Vd

0 .E Using Gauss’s Law:

0 ( . )2

W E Vd So:

0 .( ) .2

W E V d VE da By doing integration by part:

and so, V E 20 .2 v s

W E d VE da

If we take integral over all space: 20

2 allspace

W E d

Poisson’s and Laplace’s EquationE V

The fundamental equations for E:

0

. ;E

0E

2. .( )E V V Gauss’s law then says that:

2

0

V

This is known as Poisson’s equation.

In regions where there is no charge: 0 Poisson’s equation reduces to Laplace’s equation.

2 0V This is known as Laplace’s equation.