electricity phy1013s electric fields gregor leigh [email protected]

ELECTRICITY

PHY1013S

ELECTRIC

FIELDS

Gregor [email protected]

ELECTRICITY ELECTRIC FIELDSPHY1013S

2

ELECTRIC FIELDS

Learning outcomes:At the end of this chapter you should be able to…

Use a field model to explain the long-range interaction between charges.

Determine the shape and strengths of the various electric fields due to specific configurations of charge.

Calculate the forces on (and the motion of) point charges and dipoles in each of these fields.

Determine the energy necessary to rotate a dipole in a uniform electric field.


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THE CONCEPT OF A FIELD

How do some forces (gravity, electrostatics, magnetism) act at a distance?

Faraday proposed that the space itself around certain quantities (e.g. mass, charge) is “filled with influence”.It is this “altered space”, called a field, which becomes the agent acting directly on a second body in the field.

Gravitational field: A region in which a particle of mass experiences a gravitational force.

Electric field: A region in which a charged particle experiences an electrostatic force.


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I.e.:

THE GRAVITATIONAL FIELD1 2

1 on 2 2 on 1 2

m mF F G

r

Hence: 11 on 2 2 12 , toward

mF m G m

r

According to Faraday… …this is the gravitational field due to m1, written ,

……acting on m2

1mg

2 1on 2m mF m g

At Earth’s surface: Earthw mg

and 2

1

on

2

mm

Fg

m

gravitational field strength

[gEarth = 9.8 N/kg or m/s2]


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GRAVITATIONAL FIELD STRENGTH

1

112 , toward m

mg G m

r

Field strength is proportional to the mass m1

which creates the field. Larger masses cause stronger gravitational fields.

Field strength is inversely proportional to the square of the distance from m1, but never

becomes exactly zero.

Field strength does NOT depend on the mass of any other body which experiences the field. In fact, the field exists whether another mass is present to experience it or not.

Notes:


6

ELECTRIC FIELD LINES

A few imaginary “lines of force” (Faraday) are used to represent the existence of a field.

The direction of the field is given by the direction in which a test charge (positive) tends to move.

Field lines start on +ve charges and end on –ve charges.

The tangent to a field line at any point gives the direction of the electric field vector, .

Field lines never touch or cross each other.

The density of field lines is an indication of the strength of .

Field lines leave or arrive at the surface of a conductor at right angles to the surface.(What is the component of parallel to a conducting surface?)


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ELECTRIC FIELD LINES

E

E

–+

E


8

THE FIELD MODEL

Instead of applying Coulomb’s Law directly, it is often more useful to…

1. Determine the electric field at some point, due to a given configuration of “source” charge(s);

2. Calculate the force exerted by the field on an “intruder” charge at that point in the field.

– – – – – –

+ + + + + +

E

F

q


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THE ELECTRIC FIELD VECTOR, The electric field vector, , (at some point in a field) is defined as the force per unit positive charge at that point:

E

E

FEq

Units: [N/C V/m]

FEq

Magnitude of (electric field strength): E

Direction of : given by the direction of the force experienced by a (positive) test charge.

E


10

Consider two charges q1 and q2:

The field due to q1 exerts a force on q2: 2on 2 1qF q E

THE ELECTRIC FIELD VECTOR, E

And the field due to q2 exerts a force on

q1:

(Note: Fon q1 = Fon q2 …

but E1 = E2 only if q1 = q2)

q1 q2

1on 1 2qF q E

2on qF1on qF

ELECTRICITY

20

14

qE

r

20

14

qq'F

r

ELECTRIC FIELDSPHY1013S

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FIELD DUE TO A POINT CHARGE

To find the field strength at a point in the field a distance r from a point charge q, we place a test charge q' at that point.

E

According to Coulomb, the magnitude of the force between the charges is:

And therefore the electric field strength (E = F/q') is:

ELECTRICITY

20

1 ˆ4

qE r

r

20

14

qE

r


12

FIELD DUE TO A POINT CHARGE

Alternatively, in terms of the unit vector , which points straight outward from the source charge,:

The direction of is in the direction of the force on a positive test particle, i.e. away from the source charge q if E is positive, and towards q if E is negative.

E

r̂


13

VECTOR FIELD DIAGRAMS

Fields can also be represented graphically by drawing field vectors at a few select points…

The field exists everywhere – not just at the few representative points in the diagram.

The arrow indicates the strength and direction of the field at the point to which it is attached, i.e. at the tail of the vector arrow.

Although we use an arrow to represent it, the electric field vector is a point quantity – it does not “stretch” from one point to another.

Notes:

–


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ELECTRIC FIELD DUE TO MULTIPLE POINT CHARGES

The net field produced at any given point as a result of several point charges can be determined by summing the individual electric field vectors (!) at that point:

net 1 2 3 ii

E E E E E

In practice we work with the 3 simultaneous equations: net 1 2 ix x x x

E E E E

net 1 2 iy y y yE E E E

net 1 2 iz z z zE E E E


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Establish a coordinate system and draw in the charges.

Identify the point P at which you want to determine the electric field.

At P, draw each electric field vector due to each of the source charges.

Resolve each electric field vector into x-, y- and z-components.

Wherever possible, use symmetry to simplify your calculations.

ELECTRIC FIELD DUE TO MULTIPLE POINT CHARGES

Pictorial strategy:


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TYPICAL ELECTRIC FIELD STRENGTHS

Field location Field strength [N/C]

Inside a current-carrying wire

Near the Earth’s surface

Near objects charged by rubbing

Electric breakdown in air, resulting in a spark

Inside an atom

10–2

102–104

103–106

3 106

1011


17

+–

SHAPE OF THE FIELD DUE TO…

An isolated point charge:


18

–+

SHAPE OF THE FIELD DUE TO…

Two equal, unlike charges, i.e. a dipole:


19

+ +

––

ELECTRIC DIPOLES

An electric dipole is a pair of equal and opposite charges +q and –q separated by a small distance s.

Temporary dipole – formed when a neutral atom is polarised by an external charge.

Permanent dipole – atoms with differing electronegativities combine to form a polar molecule.

+ ++

HH

O


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DIPOLE MOMENT

The properties of a dipole are, essentially:

the magnitude of the charge on each pole, q;

the distance between the centres of charge, s.

We thus define the dipole moment, , as the vector: p

p

= (qs, from the negative to the positive charge)

s +q

–qp

ELECTRICITY

dipole 30

214

pE

r

dipole 30

14

pE

r

dipole 30

214

pE

y


21

FIELD DUE TO AN ELECTRIC DIPOLE

Although a dipole is neutral overall, it does create a field.y

xs

P

E

E

Q

E

E

dipoleE

dipoleE

At points along the dipole axis (at large distances from the dipole, i.e. y >> s), it can be shown that:

y In the plane around the “waist” of the dipole, (for r >> s):

+

–


22

L

The charge per unit of length is known as the linear charge density,

QL

Field “around the waist” of a uniformly charged rod:

Q

rod 220

2

14 L

QE

r r

Field due to an infinite line of charge:

line0

214

Er

FIELDS DUE TO CONTINUOUS DISTRIBUTIONS OF CHARGE

One-dimensional line of charge:


23

Ring of charge:

R

Q

The electric field on the axis of a charged ring of radius R:

ring 30 2 2

14z

zQE

z R


Pz z


24

Q

The electric field on the axis of a charged disk of radius R: disk 2 20

12z

zEz R


Disk of charge:

Pz z

R

The charge per unit of area is known as the surface charge density,

QA

A


25

plane0

( 0)2z

E z


“Infinite” plane of charge:

z

plane0

( 0)2z

E z

ELECTRICITY

sphere 20

1 ˆ4

QE r

r


26


Sphere of charge:

The electric field outside a sphere of charge (r R):

Q

R


27

Include q’s sign. If it is negative, the force q experiences is in the opposite direction to .

POINT CHARGE IN AN ELECTRIC FIELD

F qE

Millikan’s experiment:

The stationary oil drop carries three extra electronsR = 2.76 m = 920 kg/m3

= ?E

E

R, , q


28

POINT CHARGE IN AN ELECTRIC FIELD

Ink-jet printing:

Calculate y, if…

m = 1.3 10–10 kgq = –1.5 10–13 Cv0 = 18 m/s

E = 1.4 106 N/CL = 1.6 cm

Show that…

22

0

12

qEy L

mv

F qE

y

x

L

E

m, qa

m, q y

0v


29

E

sinFs

DIPOLE IN AN ELECTRIC FIELD

2 2sin sins sF F

sinpE

+q

–q

p

s/2

F

F

p E

i.e.

A dipole in a uniform electric field experiences no net force.

It does, however, experience torque about its centre of mass…

F qEbut


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DIPOLE IN AN ELECTRIC FIELD

Potential energy of a dipole :

W d

90

( )U d

( ) cosU pE

( )U p E

i.e.

U W and

U()

minimum when = 0°

set to zero when = 90°

maximum when = 180°

F

F

90

sinpE d


31

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electricity phy1013s electric fields gregor leigh [email protected]

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