Lecture notes, Chapter 21: Electric Charge and Electric Field PDF

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Chapter 21: Electric Charge and Electric Field Introduction ! Four fundamental forces of nature

(Fewer if they can be “unified”)

• Long range " gravity and electromagnetism • Short range " strong (color) and weak (flavor) nuclear force ! Electromagnetism common in our daily lives

• Solid objects held together by electric interactions • You don’t sink through your chairs because of electric interactions ! Electromagnetic interactions involve particles which have electric charge

Electric Charge (21-1) ! Electric charge is a fundamental attribute of matter ! Ancient Greeks observed amber rubbed with wool attracted other objects

• Electric derived from Greek word ηλεκτρον meaning amber ! Empirical observations (in low humidity to minimize surface adsorption of water):

• Two rubber (or amber) rods rubbed with fur repel • Two glass (plexiglas) rods rubbed with silk repel • Glass (rubbed with silk) attracts rubber rod (rubbed with fur) • The fur attracts the rubbed rubber rods • The silk attracts the rubbed glass rods ! Conclusion (from many experiments): There are two kinds of electric charge

• Positive charge on rubbed glass rod and fur • Negative charge on rubbed rubber rod and silk • Named by Benjamin Franklin Unfortunate choice: Makes electron negative. We now know that electrons are most mobile charge carriers, so current (defined as flow of positive charge) moves opposite to the actual charge carriers. Thanks Ben!

• Charges of same kind repel, different kind attract (unlike gravity) • Rubbing two different materials always charges one positive, the other negative.

V6.01 BQRS F’10

1

! Charge quantization

• Every observable amount of electric charge is always an integer multiple of a fundamental or basic unit denoted by

e

(Milliken oil drop experiment)

• The magnitude of charge of the electron or proton is the basic unit of electric charge: electron has charge

− e and proton has charge +e

• The SI unit of electric charge is called the Coulomb (C)

e = 1.602176462(63) × 10 − 19C

• Fundamental unit of charge

# Very unusual to encounter a 1C charge in electrostatic problems since it is such a huge amount of charge # In practice, the Coulomb is defined as one Ampère-second • Remarkable: all electrons have identical charge (presumed time invariant) • Atom # heavy dense core (nucleus) of protons and neutrons (zero charge) bound by the (short-ranged) strong nuclear force (order 10 #

− 15

m) m) “cloud” of light electrons bound by electric attraction (order 10 − 10

# usually equal number of protons and electrons (total charge zero) # can lose electrons " positive ion # can gain electrons " negative ion • Aside: protons and neutrons are combinations of subatomic particles known as quarks (Murray Gell-Mann, George Zweig 1963) # u quark

+ 23 e

# proton = uud = # neutron = udd =

d quark

− 13 e

+ 32 e + 23 e − 31 e = e

+ 23 e − 13 e − 13 e = 0

# hordes of other particles are known to exist:

π , K , ρ , ∆,…

# quarks never exist in isolation (only inside protons, etc.) • Antiparticles have exact opposite electric charge # Positron (antiparticle of electron) has charge # Antiproton has charge

+e

−e

• in bulk matter we will often treat charge as continuous in this course V6.01 BQRS F’10

2

! Charge conservation

• The algebraic sum of all the electric charges in any closed system is constant with time • Even in high-energy interactions in which particles and antiparticles are created or annihilated, the total net charge remains constant ! When two objects are rubbed - charge is neither created nor destroyed

• Charge just moves from one material to another: electron binding strength varies by material; rub electrons “on” or rub them “off” a surface • Positively charged objects have slightly fewer electrons than protons • Negatively charged objects have slightly more electrons than protons

V6.01 BQRS F’10

3

Coulomb’s Law (21-3) ! Charles Coulomb studied the interaction forces between charged particles in

1784 (about 100 years after Newton!) using a torsion balance ! Electrostatic force between two point charges

q1 and q 2

at rest and separated

r has magnitude 2 ⎛ 1 ⎞ q1q2 − 12 C where ε = 8.854 × 10 F =⎜ 0 ⎟ r2 πε 4 Nm 2 ⎝ 0⎠ by distance

and force is • attractive if • repulsive if

q1 and q2 q1 and q2

have opposite signs have same signs

! more convenient way to express the proportionality constant, k:

Nm 2 = 8.988 × 10 k≡ 4πε 0 C2 to make a 3D vector equation, define rˆ12 as a unit vector ( rˆ12 = 1 ) which points 1

!

9

from the source charge (1) toward the charge of interest (2) on which we wish to calculate the force

q1

rˆ12

q2

" F12 Force due to 1 acting upon 2 (watch notation)

Vector from 1 to 2 (watch notation)

! the vector force due to the source charge (1) upon the charge of interest (2)

separated by a (scalar) distance r12 may be written as:

" ⎛ 1 ⎞ q1q2 F12 = ⎜ ⎟ 2 rˆ12 πε ⎝ 4 0 ⎠ r12

“Coulomb’s Law” -valid for POINT-like PARTICLES ONLY (not rods, sheets, etc.)

! reminder: this is a vector equation.

• points from charge 1 to charge 2 if • points from charge 2 to charge 1 if

V6.01 BQRS F’10

q1q2 > 0 (same sign repulsion) q1 q2 < 0 (opposite sign attraction)

4

Some useful vectors " ! define rjA as vector from position of charge j to position of charge A ! define rˆjA as unit vector pointing from position of charge j toward " " point A of interest rjA rjA rˆjA = " = rjA rjA

q1

rˆ1A " r1A

" r4A

qA

" r2A

q2

rˆ2 A

" r3A rˆ4 A

rˆ3 A

q4

q3

iˆ, jˆ, kˆ as unit vectors pointing in the x-, y -, z -

! Define special basis vectors

directions, respectively

ˆj

" v = 3ˆi + 2 ˆj

y z

θ iˆ x

• These can be used to combine 3 equations (for the individual components) into a single vector equation • Components example:

Fx = −kx

mg cosθ + Fy = qE

Fy y

Fz = mg sin θ • is equivalent to the vector expression

" F + mg cos θ ˆj = −kx iˆ + qE ˆj + mg sin θ kˆ

V6.01 BQRS F’10

F z Fx z

x

5

Superposition of Electric Forces ! Electrostatic forces simply add like vectors

rˆ10 q0

q1

rˆ30

rˆ40

q1 q4

rˆ20

rˆ40

rˆ30

q2 q0

q rˆ10 4

q0

q2

q3

! Total force on

rˆ20

q3

4 " 1 q0 q j F0 = ∑ rˆj 0 2 4 πε r 0 j =1 j0

in the above figure is

! Example: what is total force on q3 ?

rˆ13 = ˆj

•

" F= =

q2 = 10. µC

rˆ23 = iˆ

1 ⎛ q3q1

4 πε0 ⎜⎝ r132 q3 ⎛ q1

⎜ 4 πε0 ⎝ r132

rˆ13 +

ˆj +

q3q2 r232

⎞ rˆ23 ⎟ ⎠

q2 ⎞ iˆ ⎟ 2 r23 ⎠

q3 = 10.0 µC

10.0 cm y

θ " F

14.14cm

z

x

q1 = −40.µ C

2 ⎧ ( −40. µC ) ⎛ (10. µC ) ˆ ⎫ 9 Nm ⎞ ˆ = ⎜ 8.988 ×10 + j i 10.0 C µ ( ) ⎨ 2 ⎟ 2 2 ⎬ C ( 0.100m ) ⎭ ⎝ ⎠ ⎩ ( 0.1414m )

= ( 90iˆ − 180 ˆj ) N • Equivalent to giving components: • Or magnitude: • And angle

Fx = 90. N, Fy = − 1.8 × 102 N, Fz = 0

" F = Fx2 + Fy2 + Fz2 = 2.0 × 102 N

tan θ = Fy / Fx = −2.0 ⇒ θ = tan −1( −2.) = −63 ° OR ⇒ θ = tan −1( −2.) + 180 ° = 117 °

(Here, Fy is negative so V6.01 BQRS F’10

θ = − 63° is correct.) 6

• Note on significant figures: results are never more precise than input data

q2

due to q1 ?

r12 = r12 =

( −0.100)

! Example: what is force on

"

• Displacement vector from 1"2: r12 = " 2 • Distance: • Unit displacement vector: rˆ12 • the force is then

" F12 =

( −0.100 iˆ + 0.141 ˆj) m

+( 0.141) m = 0.173 m 2

= − 0.578 iˆ + 0.816 ˆj

2 ⎛ 1 q2 q1 9 Nm ⎞ (10. µC )( −40. µC ) rˆ = ⎜ 8.988 × 10 rˆ12 2 12 2 ⎟ 2 4πε 0 r12 C ⎠ ( 0.173 m ) ⎝

= −120. ( −0.578iˆ + 0.816 ˆj ) N = ( 70. iˆ − 98. ˆj ) N "

"

"

"

• (Force due to q3, F32 could be found similarly, then added to get F2 = F12 + F32 )

V6.01 BQRS F’10

7

Electric Field (21-4) ! the total electric force on any small test charge

!

is proportional to q 0

⎧ N 1 qi ⎫ 1 q 0q i rˆi 0 = q0 ⎨ rˆ 2 2 i0 ⎬ 4 ⎩ i =1 4πε 0 ri 0 ⎭ i =1 πε 0 ri 0 " the part multiplying q0 is called the electric field E " " Ftot = q0 E " " electric field E (r ) " Ftot =

!

q0

N

∑

∑

•

points in the direction in which a positive charge would be pushed

•

has a vector value at every point in space (is a vector field) (even if there is no charge there to ‘feel’ it)

•

is created by electric charges

•

is the means by which charged particles interact

•

has units of force per unit charge N/C = (V/m)

•

later, we’ll see that

" " E (r ) contains energy and can carry momentum!

! Coulomb’s law revisited: electrostatic force of charge A on B can be viewed as: •

Charge q A

creates an electric field

" " EA ( r ) which permeates the

surrounding space (modifies the properties of the space around it) •

A charge

qB

located at position

exerted by the electric field of •

" rAB

" " " feels a force FAB = q B E A ( rAB )

qA

In other words: The electric force on a charged body is exerted by the electric field created by other charged bodies

! For a single point charge q at rest:

" " 1 q E(r ) = rˆ 4πε 0 r2

•

rˆ points from charge q to point of interest " for q > 0, E points radially away from charge " q < 0, E points radially toward charge for

•

magnitude falls off with distance r from q

• •

V6.01 BQRS F’10

8

Conductors, Insulators, and Induced Charges (21-2) ! Insulators (dielectrics) don't let electrons move around " if they become

charged, they hold the charges in place ! Conductors (metals) allow charge to flow, so charge spreads out, repelled from

one another"conductors easily polarized by being near a charged object ! Conductors can be discharged by grounding them

• grounding: connecting the conductor to the earth (ground) with a conducting path (e.g. wire, professor, ...) • the earth is a huge conductor that it can be considered as a bottomless source or sink of electrons • if a conductor is not near a charged object and is connected to the ground, excess charges repel each other and flow off of the conductor until it is electrically neutral • if conductor is near a charged object and is connected to ground, excess charges flow so as to balance electrostatic attraction/repulsion due to charged object and each other ! Two ways a conductor can be charged:

• by touching or rubbing it with an already charged object • by induction without ever touching it with a charged object! ! Charging a conductor by induction

• Bring charged rod near conductor " conductor becomes polarized

V6.01 BQRS F’10

9

Not all free electrons move to right since attraction of the positive charge built up on the left becomes large enough to offset the repulsion due to the rod. • Allow repelled charges to flow away to ground (or to another conductor) • Disconnect wire to ground and remove charged rod

! Electric forces on uncharged objects also exist !

• At the molecular level, uncharged insulators are often polarizable • The distorted charge distribution leads to a net force (by superposition, see 21.3) of attraction • Can be very important in chemistry • If neutral object touches charged one and becomes charged, then the force becomes repulsive... can get recursive!

V6.01 BQRS F’10

10

Electric Field of a Charge Distribution (21.5) ! In practice, we deal with charges which are distributed over space ! We need to know how to compute the electric field of such charge distributions ! Exploit the principle of superposition: •

The electric field of several charges is the vector sum of the separate electric fields of each charge

•

Superposition: each charge acts independently of the others $ Note: you have seen that gravity acts like this, too, but not so the

nuclear strong and weak forces ! Total electric field (or total force on a test charge) can be calculated by summing

electric fields (or forces) due to the infinitesimal pieces of the charge distribution

"

! Electric field at some point rP •

set of discrete point charges:

•

continuous charge distribution:

" " E ( rP ) = " " E ( rP ) =

1 4 πε0 1 4 πε0

N

qj

∑r j=1

∫

2 jP

rˆjP

dQ rˆdQ P 2 rdQ P

! Kinds of charge distributions: •

Line charge

•

Surface charge

•

Volume charge

V6.01 BQRS F’10

λ (Units: C/m) σ (Units: C/m 2 ) ρ (Units: C/m3 )

11

! Example: field of an electric dipole •

y

What is electric field at point A at (x,0,0)?

q2 = q d

q1 = −q, q2 = q

d " r1A = xiˆ + ˆj ⇒ r1A = 2

" ⎞ 1 ⎛ q1 q2 ˆ ˆ + EA = r r ⎜ 1A 2A ⎟ 4πε 0 ⎝ r12A r22A ⎠ ⎛ ˆ + d jˆ xi −q q 1 ⎜ 2 = + 4πε 0 ⎜⎜ x 2 + ( d / 2) 2 x 2 + (d / 2)2 x 2 + ( d / 2) 2 ⎝

(

V6.01 BQRS F’10

A x

d xiˆ + ˆj 2 x2 + (d / 2)2 , rˆ1A = x 2 + (d / 2) 2

" d r2A = xiˆ − ˆj ⇒ r2A = x2 + ( d / 2)2 , rˆ2A = 2

⎛ −q d jˆ 1 ⎜ = 4πε 0 ⎜⎜ x 2 + (d / 2) 2 ⎝

x q1 = −q

d xiˆ − ˆj 2 x 2 + (d / 2) 2

d ⎞ xiˆ − jˆ ⎟ 2 ⎟ x 2 + (d / 2)2 ⎟ ⎠

⎞ ⎟ 3/2 ⎟ ⎟ ⎠

)

12

! Example: find

" E

at

(0, 0, b)

z

on axis of uniformly

charged ring in xy-plane with radius a, charge Q •

•

charge per unit length on ring

λ=

" rdQ P

b

Q 2π a

Q

y a

infinitesimal element of charge

φ

dQ

x

Q dQ = λ dl = λ a dφ = dφ 2π •

P

vector from charge element,dQ , to selected point, P

" r( dQ ) P = −a cos φ iˆ − a sin φ jˆ + b kˆ " r( dQ ) P = r( dQ) P = a 2 cos 2 φ + a 2 sin 2 φ + b 2 = a 2 + b 2 " − a cos φ iˆ − a sin φ ˆj + b kˆ r( dQ ) P rˆ( dQ ) P = = r( dQ ) P a 2 + b2

(

•

)

Electric field at point P

" EP =

1

∫

dQ

4 πε0 r(2dQ) P

rˆdQ P

(

2π − a cos φ ˆi − a sin φ ˆj + b kˆ 1 ⎛Q ⎞ = ⎜ ⎟ dφ 3/ 2 4 πε0 ⎝ 2 π ⎠ 0 a2 + b2

∫

=

(

Qb

1

(

4 πε0 a 2 + b

)

2 3/2

)

)

kˆ

•

As expected, x and y components must be zero by symmetry

•

For large

V6.01 BQRS F’10

b # a , field looks like that of a point charge " 1 1 Q ˆ Qb ˆ = sgn( ) EP → k b k, 4πε 0 (| b |) 3 4πε 0 b 2

b #a

13

! Example: find

" E

at distance b away, along the

midplane of charge Q uniformly spread over line (wire) along z-axis from

z = −a

to

z=a

•

charge per unit length on wire

•

infinitesimal element of charge

dQ = λ dz =

Q λ= 2a

z

kˆ a dQ z

" rdQ P

b

ρ

P ρˆ

−a

Q dz 2a

•

Use cylindrical coordinates: ρ is the distance from the z-axis " " Let r be shorthand for r(dQ ) P

•

vector from charge element to selected point P

" r = b ρˆ − z kˆ ⇒ r = b2 + z 2

b ρˆ − z kˆ ) ( rˆ = b2 + z 2

•

Electric field at point P due to one piece of charge, dQ

" 1 dQ 1 ⎛ Q ⎞ dz = dEP = r ˆ ⎜ ⎟ 2 rˆ 2 4πε 0 r 4πε 0 ⎝ 2a ⎠ r •

Total Electric field at point P

" EP = •

1 4πε 0

∫

( ∫ (

)

a b ρˆ − z kˆ dQ 1 ⎛Q⎞ rˆ = ⎜ ⎟ dz 2 3/ 2 2 4πε 0 ⎝ 2 a ⎠ −a r2 b +z

)

Integrals: a

∫

−a

z dz

(b

2

+z

)

2 3/2

= 0 (integral of odd function over even region z → − z )

$ Note how z-component was “expected” to be zero by symmetry

V6.01 BQRS F’10

14

a

∫ (b

−a

b dz 2

+z

)

2 3/2

⎧ ⎪ z = b tan θ ⎪ ⎪ ⎨so ⎪ b dz ⎪ ⎪∫ ⎩ (b + z ) 2

•

2

2a b a +b 2

b dθ

⇒ dz=

cos θ

−1 tan

(a/b)

∫

= 3/ 2

cos θ dθ

Final result:

)

b

3/ 2

2

+z

sin θ b

b

−1 − tan

=

z = b tan θ )

(substitute

2

(b

and 2

a

−a

=

2

tan

−1

=

3

3

co s θ

(a /b )

2

a

b

a +b

= − tan

−1

( a / b)

(a /b )

" EP =

1

Q

4πε 0 b a2 + b2

b #a

•

Far from short wire

•

For long wire

•

For any point in xy plane?

2

2

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

ρˆ

field of point charge

" EP $

Q ρˆ 4πε 0 b2 1

a # b (keeping charge density λ fixed) " 1 λ (falls only as 1/distance !!) EP = ρˆ 2πε 0 b

$ Field points radially away from wire ...