Physics 121: Electricity & Magnetism – Lecture 3

Electric Field

Dale E. GaryWenda Cao

NJIT Physics Department

September 18, 2007

Electric Force and Field Force

What? -- Action on a distance

How? – Electric Field

Why? – Field Force Where? – in the

space surrounding charges

September 18, 2007

Fields Scalar Fields:

Temperature – T(r) Pressure – P(r) Potential energy –

U(r) Vector Fields:

Velocity field – Gravitational field – Electric field – Magnetic field –

)(rv

)(rg

)(rE

)(rB

September 18, 2007

Vector Field Due to Gravity When you consider the

force of Earth’s gravity in space, it points everywhere in the direction of the center of the Earth. But remember that the strength is:

This is an example of an inverse-square force (proportional to the inverse square of the distance).

rr

MmGF ˆ

2

m

M

September 18, 2007

Idea of Test Mass Notice that the actual

amount of force depends on the mass, m:

It is convenient to ask what is the force per unit mass. The idea is to imagine putting a unit test mass near the Earth, and observe the effect on it:

g(r) is the “gravitational field.”

rr

GMmF ˆ

2

rrgrr

GM

m

Fˆ)(ˆ

2

September 18, 2007

Electric Field Electric field is said to exist

in the region of space around a charged object: the source charge.

Concept of test charge: Small and positive Does not affect charge

distribution Electric field:

Existence of an electric field is a property of its source;

Presence of test charge is not necessary for the field to exist;

0q

FE

+ + +

+++ +

+

+

September 18, 2007

Electric Field

1. A test charge of +3 µC is at a point P where an external electric field is directed to the right and has a magnitude of 4×106 N/C. If the test charge is replaced with another test charge of –3 µC, what happens to the external electric field at P ?

A. It is unaffected. B. It reverses direction. C. It changes in a way that cannot be

determined.

September 18, 2007

Electric Field

0q

FE

Magnitude: E=F/q0

Direction: is that of the force that acts on the positive test charge

SI unit: N/C

Situation Value

Inside a copper wire of household circuits 10-2 N/C

Near a charged comb 103 N/C

Inside a TV picture tube 105 N/C

Near the charged drum of a photocopier 105 N/C

Electric breakdown across an air gap 3×106 N/C

At the electron’s orbit in a hydrogen atom 5×1011 N/C

On the suface of a Uranium nucleus 3×1021 N/C

September 18, 2007

2. Which diagram could be considered to show the correct electric force on a positive test charge due to a point charge?

A. B.

C. D. E.

September 18, 2007

Electric Field due to a Point Charge Q

rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00

Direction is radial: outward for +|Q| inward for -|Q| Magnitude: constant on any

spherical shell Flux through any shell enclosing Q

is the same: EAAA = EBAB

A

B

Q

q0

r

September 18, 2007

Electric Field due to a group of individual charge

nFFFF 002010 ...

n

n

EEE

q

F

q

F

q

F

q

FE

...

...

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0

September 18, 2007

Start with

If d << z, then,

So

Example: Electric Field of a Dipole

])2

1()2

1[(4

)2/(4

1

)2/(4

1

4

1

4

1

222

0

20

20

20

20

z

d

z

d

z

q

dz

q

dz

q

r

q

r

qEEE

30

30

20

2

1

2

12

4

z

pE

qdp

z

qd

z

d

z

qE

z

d

z

d

z

d

z

d

z

d 2...)]

)!1(2

21(...)

)!1(2

21[(])

21()

21[( 22

E ~ 1/z3

E =>0 as d =>0 Valid for “far field”

September 18, 2007

Find an expression for dq: dq = λdl for a line distribution dq = σdA for a surface

distribution dq = ρdV for a volume

distribution Represent field contributions at P

due to point charges dq located in the distribution. Use symmetry,

Add up (integrate the contributions) over the whole distribution, varying the displacement as needed,

Electric Field of a Continuous Charge Distribution

rr

qE ˆ

4

12

0

ii

i

i rr

qE ˆ

4

12

0

rr

dqr

r

qE

ii

i

i

qˆ

4

1ˆlim

4

12

020

0

r

r

dqEd ˆ

4 20

dEE

September 18, 2007

A rod of length l has a uniform positive charge per unit length λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end.

Start with

then,

So

Example: Electric Field Due to a Charged Rod

20

20 4

1

4

1

x

dx

x

dqdE

dxdq

al

a

al

a

al

a xx

dx

x

dxE

1

444 02

02

0

)(4

11

4

1

00 ala

Q

alal

QE

Finalize l => 0 ? a >> l ?

September 18, 2007

Electric Field Lines The electric field vector is tangent

to the electric field line at each point. The line has a direction, indicated by an arrowhead, that is the same as that of the electric field vector. The direction of the line is that of the force on a positive test charge placed in the field.

The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region. Thus, the field lines are close together where the electric field is strong and far apart where the field is weak.

September 18, 2007

Electric Field Lines The lines must begin on a positive

charge and terminate on a negative charge. In the case of an excess of one type of charge, some lines will begin or end infinitely far away.

The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.

No two field lines can cross.

September 18, 2007

Electric Field3. Rank the magnitudes E

of the electric field at points A, B, and C shown in the figure.

A) EC>EB>EA

B) EB>EC>EA

C) EA>EC>EB

D) EB>EA>EC

E) EA>EB>EC

.C

.A

.B

September 18, 2007

Motion of a Charged Particle in a Uniform Electric FieldEqF

m

Eqa

amEqF

If the electric field E is uniform (magnitude and direction), the electric force F on the particle is constant.

If the particle has a positive charge, its acceleration a and electric force F are in the direction of the electric field E.

If the particle has a negative charge, its acceleration a and electric force F are in the direction opposite the electric field E.

September 18, 2007

Start with

Then and

So

A Dipole in an Electric Field

sinsin)(sin FdxdFFx

qdp qEF

p E

sin|| pE

September 18, 2007

Start with

Since

Choose at So

A Dipole in an Electric Field

)cos(cos]cos[

sinsin

fi

if

pEpE

dpEdpEdUU

f

i

f

i

f

i

f

i

ddW

cospEU

U p E

0iU 90i

September 18, 2007

4. In which configuration, the potential energy of the dipole is the greatest?

E

a b c

d e

September 18, 2007

Summary Electric field E at any point is defined in terms of the electric force F that acts on a small

positive test charge placed at that point divided by the magnitude q0 of the test charge:

Electric field lines provide a means for visualizing the direction and magnitude of electric fields. The electric field vector at any point is tangent to a field line through that point. The density of field lines in any region is proportional to the magnitude of the electric field in that region.

Field lines originate on positive charge and terminate on negative charge. Field due to a point charge:

The direction is away from the point charge if the charge is positive and toward it if the charge is negative.

Field due to an electric dipole:

Field due to a continuous charge distribution: treat charge elements as point charges and then summing via inegration, the electric field vectors produced by all the charge elements.

Force on a point charge in an electric field:

Dipole in an electric field: The field exerts a torque on the dipole

The dipole has a potential energy U associated with its orientation in the field

rr

Q

q

FE ˆ

4

12

00

p E

U p E

302

1

z

pE

0q

FE

EqF