Chapter 23 Electric Charge and Electric Fields

• What is a field?

• Why have them?

• What causes fields?

Field Type Caused By

gravity mass

electric charge

magnetic moving charge

Electric Charge• Types:

– Positive• Glass rubbed with silk • Missing electrons

– Negative• Rubber/Plastic rubbed with fur• Extra electrons

• Arbitrary choice – convention attributed to ?

• Units: amount of charge is measured in [Coulombs]

• Empirical Observations:– Like charges repel– Unlike charges attract

Charge in the Atom

• Protons (+)

• Electrons (-)

• Ions

• Polar Molecules

Charge Properties• Conservation

– Charge is not created or destroyed, only transferred.

– The net amount of electric charge produced in any process is zero.

• Quantization– The smallest unit of charge is that on an electron or

proton. (e = 1.6 x 10-19 C)

• It is impossible to have less charge than this

• It is possible to have integer multiples of this charge

Q Ne

Conductors and Insulators

• Conductor transfers charge on contact• Insulator does not transfer charge on contact• Semiconductor might transfer charge on contact

Charge Transfer Processes

• Conduction

• Polarization

• Induction

The Electroscope

Coulomb’s Law• Empirical Observations

• Formal Statement

1 2F q q

2

1F

r

Direction of the force is along the line joining the two charges

1 212 212

21

kq qˆF r

r

Active Figure 23.7

(SLIDESHOW MODE ONLY)

Coulomb’s Law Example

• What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m

Hydrogen Atom Example

• The electrical force between the electron and proton is found from Coulomb’s law– Fe = keq1q2 / r2 = 8.2 x 108 N

• This can be compared to the gravitational force between the electron and the proton– Fg = Gmemp / r2 = 3.6 x 10-47 N

Subscript Convention

1 212 212

21

kq qˆF r

r

12 1 2F force on charge q due to charge q

21 2 1r distance from charge q to charge q

21 2 1r̂ unit vector oriented from charge q to charge q

+q1+q2

21r̂ 12F

21r

More Coulomb’s Law

+q1+q2

21r̂

Coulomb’s constant:2 2

9 92 2

o

N m N m 1k 8.988x10 9.0x10

C C 4

permittivity of free space:2

12o 2

1 C8.85x10

4 k N m

1 212 212

21

kq qˆF r

r

12F

21r

21 21r r

2121

21

rr̂

r

Charge polarity: Same sign Force is rightOpposite sign Force is Left

Electrostatics --- Charges must be at rest!

Superposition of Forces

0 01 02 03F F F F ....

0 1 0 2 0 30 10 20 302 2 2

10 20 30

kq q kq q kq qˆ ˆ ˆF r r r ....

r r r

N31 2 i

0 0 10 20 30 0 i02 2 2 2i 110 20 30 i0

qq q qˆ ˆ ˆ ˆF kq r r r .... kq r

r r r r

+Q3

+Q2

+Q1

+Q0

01F

03F

02F

10r

20r

30r

Coulomb’s Law Example

• Q = 6.0 mC• L = 0.10 m• What is the magnitude

and direction of the net force on one of the charges?

F1

+

F2

F3

Q

x

y

L

++

+

L

QQ

Q

Zero Resultant Force, Example

• Where is the resultant force equal to zero?– The magnitudes of the

individual forces will be equal

– Directions will be opposite

• Will result in a quadratic

• Choose the root that gives the forces in opposite directions

Electrical Force with Other Forces, Example

• The spheres are in equilibrium

• Since they are separated, they exert a repulsive force on each other– Charges are like charges

• Proceed as usual with equilibrium problems, noting one force is an electrical force

Electrical Force with Other Forces, Example cont.

• The free body diagram includes the components of the tension, the electrical force, and the weight

• Solve for |q| • You cannot

determine the sign of q, only that they both have same sign

The Electric Field• Charge particles create forces on each other without ever

coming into contact. » “action at a distance”

• A charge creates in space the ability to exert a force on a second very small charge. This ability exists even if the second charge is not present.

• We call this ability to exert a force at a distance a “field”• In general, a field is defined:

• The Electric Field is then:

test quantity 0

ForceField lim

test quantity

q 0

FE lim

q

N

C

Why in the limit?

Electric Field near a Point Charge

oq 0o

FE lim

q

o2

kq qˆF r

r

o

o2

2q 0o

kqqr̂ kqr ˆE lim r

q r

+Q -Q

Electric Field Vectors

Electric Field Lines

Active Figure 23.13

(SLIDESHOW MODE ONLY)

Rules for Drawing Field Lines• The electric field, , is tangent to the field lines.• The number of lines leaving/entering a charge is

proportional to the charge.• The number of lines passing through a unit area normal

to the lines is proportional to the strength of the field in that region.

• Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention.

• No two field lines can cross.

E

# of electric field linesE

Area

Electric Field Lines, General

• The density of lines through surface A is greater than through surface B

• The magnitude of the electric field is greater on surface A than B

• The lines at different locations point in different directions– This indicates the field is

non-uniform

Example Field Lines

+ + + + +

Dipole Line Charge

Q dq

dx

Linear Charge Density:

For a continuous linear charge distribution,

Active Figure 23.24

(SLIDESHOW MODE ONLY)

More Field Lines

Q dq

A dA Surface Charge Density:

Q dq

V dV Volume Charge Density:

Superposition of Fields

0 01 02 03E E E E ....

31 20 10 20 302 2 2

10 20 30

kqkq kqˆ ˆ ˆE r r r ....

r r r

N31 2 i

0 10 20 30 i02 2 2 2i 110 20 30 i0

qq q qˆ ˆ ˆ ˆE k r r r .... k r

r r r r

+q3

+q2

+q1

01E

03E

02E

10r

20r

30r0

Superposition Example

• Find the electric field due to q1, E1

• Find the electric field due to q2, E2

• E = E1 + E2

– Remember, the fields add as vectors

– The direction of the individual fields is the direction of the force on a positive test charge

Electric Field of a Dipole (ex. 23.6)

p

2a-q +q

E

y

E

E E E

22

kqE E

y a

x xE E E 2E cos

32 2 2 22 2 2

kq a kpE 2

y a y a y a

y 2a3

kpE

yp 2aQ

P23.19

Three point charges are arranged as shown in Figure P23.19.

(a) Find the vector electric field that the 6.00-nC and –3.00-nC charges together create at the origin.

(b) (b) Find the vector force on the 5.00-nC charge.

Figure P23.19

P23.52

Three point charges are aligned along the x axis as shown in Figure P23.52. Find the electric field at

(a) the position (2.00, 0) and

(b) the position (0, 2.00).

Figure P23.52

9 2 2 9

1 2 2

8.99 10 N m C 4.00 10 Cˆˆ

2.50 m

ˆ5.75 N C

ek q

r

E r i

i

9 2 2 9

2 2 2

8.99 10 N m C 5.00 10 Cˆ ˆˆ 11.2 N C

2.00 mek q

r

E r i i

9 2 2 9

3 2

8.99 10 N m C 3.00 10 Cˆ ˆ18.7 N C

1.20 m

E i i

1 2 3 24.2 N CR E E E E

FIG. P23.52(a)

in +x direction.

1 2ˆ ˆˆ 8.46 N C 0.243 0.970ek q

r E r i j

2 2

3 2

1 3 1 2 3

ˆˆ 11.2 N C

ˆ ˆˆ 5.81 N C 0.371 +0.928

ˆ ˆ4.21 N C 8.43 N C

e

e

x x x y y y y

k q

rk q

r

E E E E E E E

E r j

E r i j

i j

9.42 N CRE 63.4 above axisx

P23.19

9 91 3

1 2 21

8.99 10 3.00 10ˆ ˆ ˆ2.70 10 N C

0.100

ek q

r

E j j j

9 92 2

2 2 22

2 32 1

8.99 10 6.00 10ˆ ˆ ˆ5.99 10 N C

0.300

ˆ ˆ5.99 10 N C 2.70 10 N C

ek q

r

E i i i

E E E i j

9 ˆ ˆ5.00 10 C 599 2700 N Cq F E i j

6 6ˆ ˆ ˆ ˆ3.00 10 13.5 10 N 3.00 13.5 N F i j i j

(a)

(b)

Continuous Charge Distributions

Ni

0 i02i 1 i0

qˆE k r

r

0 2all charge

dqˆE k r

r

Discrete charges Continuous charge distribution

0 2

kqˆE r

r

Single charge

0 2

kdqˆdE r

r

Single piece of a charge distribution

+Q3

+Q2

+Q1

01E

03E

02E

10r

20r

30r 0 0

++

++

0dE

dq

Finding dq

Q dq

dx

Line charge dq dx dq Rd

Cartesian Polar

Surface chargeQ dq

A dA dq dxdy dq rdrd

Volume chargeQ dq

V dV dq dxdydz dq rdrd dz

2dq r sin drd d

2

kdqˆdE r

r

Example – Infinitely Long Line of Charge

+

+

+

+

dE

+

+

+

dy

x

dq dy

y

2 2 2r x y

2

kdqˆdE r

r

xdE

ydE

+

y-components cancel by symmetry

x 2

kdqdE cos

r

2 2 2 2

k dy xdE

x y x y

3 2

2 2 2

dy 2 2kE k x k x

x xx y

Example – Charged Ring (ex 23.8)

dE

x

dq ds ad

2 2 2r x a

xdE

ydE

a

d

+

+

+

+ +

+

+

2

kdqˆdE r

r

y-components cancel by symmetry

x 2

kdqdE cos

r

2 2 2 2

k ad xdE

x a x a

2

3 3 32 2 2 2 2 202 2 2

k xa k xa kQxE d 2

x a x a x a

Check a Limiting Case

3

2 2 2

kQxE

x R

When: x R

The charged ring must look like a point source.

3 3 3 2

2 2 22 22

2

kQx kQx kQx kQE

x xx R R

x 1x

0

Uniformly Charged Disk (ex. 23.9)

dE

x

3

2 2 2

kQxE

x r

r

3

2 2 2

kxdqdE

x r

dq dA rdrd 2 rdr

3

2 2 2

kx 2 rdrdE

x r

R

2 2

2

R R x R

3 3 32 2 2 20 02 2 2x

kx 2 rdr 2rdr duE kx kx

x r x r u

2 2

2 2

2

2

x R1

x R 3 22

2 2 2 2 2x

x

u 1 1 xkx u du kx 2kx k 2 1

1 x R x x R2

Binomial Expansion Theorem

n 2n n 11 x 1 nx x ...

2!

x 1 Quadratic terms and higher are small

n1 x 1 nx

Two Important Limiting Cases

2 2o o

x 1E k 2 1 k 2 2

4 2x R

Large Charged Plate: R x

dE

x

rR

Very Far From the Charged Plate: x R

12 2

22 2 2

2

x x RE k 2 1 k 2 1 k 2 1 1

xx R Rx 1

x

2 2 2

2 2 2 2

1 R 1 R k R kQk 2 1 1 k 2

2 x 2 x x x

Parallel Plate Capacitor

-Q

+Q

o

E2

o

E

E 0

E 0

Motion of Charged Particles in a Uniform Electric Field

-e

-Q +Q

q 0

FE lim

q

F qE ma

qEa

m

2 20 0v v 2a x x

x x

e Ev 2a x 2 x

m

x

Example

• A proton accelerates from rest in a uniform electric field of 500 N/C. At some time later, its speed is 2.50 x 106 m/s.– Find the acceleration of the

proton.– How long does it take for the

proton to reach this speed?– How far has it moved in this time?– What is the kinetic energy?

e

-Q +Q

x

Motion of Charged Particles in a Uniform Electric Field

-e

+Q

-Q

q 0

FE lim

q

F qE ma

y

e Ea

m

x x0 x x0v v a t v

y y0 y

e Ev v a t t

m

x0v

2

2 2 2x y x0

e E tv v v v

m

v

1

x0

e E ttan

mv

Active Figure 23.26

(SLIDESHOW MODE ONLY)

Motion of Charged Particles in a Uniform Electric Field

-e-e

-Q +Q

+Q

-Qx

Phosphor Screen

This device is known as a cathode ray tube (CRT)

Summary

Coulomb’s Law1 2

12 21221

kq qˆF r

r

oq 0o

FE lim

q

Electric Field

Point Charges:

These can be used to find the fields in the vicinity of continuous charge distributions:

2kE

x

Line of Charge:

o

E2

Charged Plate:

32 2 2

kpE

r a

Dipole:p

+

++++

dE

x

rR