Electric Charge, Force, and Fields: Understanding Electricity, Slides of Physics

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20. Electric Charge, Force, & Fields

20.1. Electric Charge

2 kinds of charges: + &



.

Total charge = algebraic sum of all charges.Like charges repel.

Opposite charges attract.

i

i

Q

q

^

3 d x

r

All electrons have charge



e.

All protons have charge +

e

.

19

e

C



= elementary charge

Theory (standard model) : basic unit of charge (carried by quark) = 1/

e

.

Quark confinement



no free quark can be observed.



Smallest observable charge is

e

.

1

st

measured by Millikan on oil drops.

Conservation of charge: total charge in a closed region is always the same.

Conceptual Example 20.1. Gravity & Electric Force

The electric force is far stronger than the gravitational force,yet gravity is much more obvious in everyday life.Why?

Only 1 kind of gravitational “charge” 

forces from different parts of a source tend to reinforce.

2 kinds of electric charges 

forces from different parts of a neutral source tend to cancel out.

Making the Connection

Compare the magnitudes of the electric & gravitationalforces between an electron & a proton.

2 e^

p

g

m m

F

G

r

2 2

E

k e

F

r



2

E g^

e^

p

F

k e

F

G m m 

^

^

^

^

^

^

^

2

9

2

2

19

11

2

2

31

27

/^

/^

N

m

C

C

N

m

kg

kg

kg



^

^



^

^

39

Example 20.2. Raindrops

Charged raindrops are responsible for thunderstorms.Two drops with equal charge

q

are on the

x

-axis at

x

=



a

.

Find the electric force on a 3

rd

drop with charge

Q

at any point on the

y

-axis.

^

^

2

2

cos

, sin

cos

, sin

k q Q

k q Q

r

r

2

0 , sin

k q Q

r

3

k q Q y

r

j

^

2

2

k q Q y a

y

j

sin

y r

y

x

x 1

=



a

x 2

= a

q

q

Q

y

r^

r F

1

F

2





1

2

F

F

F

2

2

r

a

y

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20.3. The Electric Field

Electric field

E

at

r

= Electric force on unit point charge at

r

.

1 q  E

F

F

= electric force on point charge

q

.

Gravitational field

Electric field

Implicit assumption: q

doesn’t disturb

E

.

0

lim q^

q



 E

F

Rigorous definition:

[^

E

] = N / C

= V / m

V = Volt

g

=

F

/m

E

=

F

/q

The Field of a Point Charge

Field vectors for a negative point charge.

1^ test q  E

F

2

k qr 

r

Field at

r

from point charge

q

:

2

test

test

k q q

q

r

E

r

20.4. Fields of Charge Distributions

Superposition principle



i

i





E

E

2

i

i

i^

i k qr



r

(Discrete sources)(Point charges)

Example 20.5. Modeling a Molecule

A molecule is modeled as a positive charge

q

at

x = a

,

and a negative charge



q

at

x =

a

.

Evaluate the electric field on the

y

-axis.

Find an approximate expression valid at large distances (

y

a

).

^

2

2

x

k

q

a

k q

a

E

r

r

r

r

2

2

y

k

q

y

k q

y

E

r

r

r

r

3

2

k qar

^

3/

2

2

k qa

a

y

^3

k qay

( y

a

)

y

x

x 1

=



a

x 2

= a



q

q

Q = 1 y

r^

r

E E^21 



E

Dipole (



q

with separation

d

):

3 q d

E

r



for

r

d = 2a

Typical of neutral, non-spherical, charge distributions (

d

~ size ).

Dipole moment :

p

=

q

d

.

On perpendicular bisector:

3

k py

E

i

On dipole axis:

3

2

k px

E

i

(Prob

習題

y

x

x 1

=



d/

x 2

= d/

 q

q

Q = 1

y

r^

r

E E^21 



E

d

= vector from



q to +q

3 k y

p

p

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Example 20.6. Charged Ring

A ring of radius

a

carries a uniformly distributed charge

Q

.

Find

E

at any point on the axis of the ring.

By symmetry,

E

has only axial (x-) component.

x^

x

Ring

E

dE

2

Ring

k dq

x

r

r 

^

2

2

Ring

k x

dq

a

x

^

2

2

k Q x a

x

^

2

2

k Q x a

x

E

i^

On axis of uniformly charged ring

Example 20.7. Power Line

A long electric power line running along the

x

-axis

carries a uniform charge density



[C/m].

Find

E

on the

y

-axis, assuming the wire to be infinitely long.

By symmetry,

E

has only

y

• component.

y^

y

Line

E

dE

2

Line

k dq

y

r

r 

k

 

E

ρ

Perpendicular to an infinite wire

^

2

2 dx

k y

y

x

 

2

Line

k

dx

y

r

r

^2

2

2

x

k y

y

y

x

 

2

2

k y

y

y

k y



y

x

dq

P x

y

r^

r dE

dE

dq

dE

y

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Example 20.8. Electrostatic Analyzer

Two curved metal plates establish a field of strength

E

=

E

0

( b/r

),

where

E

0

&

b

are constants.

E

points toward the center of curvature, &

r

is the distance to the center.

Find speed

v

with which a proton entering vertically from below will leave

the device moving horizontally.

For a uniform circular motion:

2

0

v

b

m

e E

r

r

0

e

v

E b m



Too fast, hitsouter wall

Too slow, hitsinner wall

Dipoles in Electric Fields

Uniform

E

:

^

q

q

F

E

E

^

^

^

ˆ^

d

d

q

q

τ

p

E

p

E

^0

Total force:Torque about center of dipole:

d q 

p

E

τ

p

E

Work done by

E

to rotate dipole :

f i

W

r d

 

F t

^

sin

sin

f i

d

W

qE

qE

d

 

t^

// tangent

sin f i

p E

d

 

^

cos

cos

f^

i

p E

Potential energy ofdipole in

E

(

i^

=



/2)

cos

f

U

W

p E

p E

( U = 0 for

p



E

)

= dipole moment

ˆ

q d  p

p

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