EX. Potential for uniformly charged thin ring

Qr z

R

φddQ

?)( =ZV 0

20

0

200

0

0

2 2

, 2 2

2

2

Q

k dQ Q QV dQ Rd dr Rk Q drk Q dr

k Qrk QR z

π

π

ϕ ϕπ π

ϕπ

ϕπ

= = ⋅ =∫

= ⋅∫

= ∫

=

=+

EX. Potential for uniformly charged spherical shell

0

2

2 2

2 20 00 0

,

sin sin ,4 4 4

sin ( )4

0 0constant in shpere

f i

k dQVrQ Q QdQ dA R d d d dR R

k Q k QV R d d r Rr r

r R E V V E dsV

π π

θ θ ϕ θ θ ϕπ π π

θ θ ϕπ

=

= = =

= = >

< ⇒ = ⇒ − = − ⋅ =

⇒ =

∫

∫ ∫

∫r r

r

Q?=V

R dQ

E

r

V

rR R

A

B

C

pr Consider the electric dipole shown in the figure.We will determine the electric potential created at point P by the two charges of the dipole using superposition.Point P is at a distance from the

V

r

( ) ( )( ) ( )

( ) ( ) ( ) ( )

center O of the dipole.Line OP makes an angle with the dipole axis

14 4o o

r rq q qV V Vr r r r

θ

πε πε− +

+ −+ − − +

⎛ ⎞ −= + = − =⎜ ⎟⎜ ⎟

⎝ ⎠

(24 - 6)

Example :

Potential due to an electric dipole

A

B

C

pr( ) ( )

2( ) ( ) 2 2

We assume that where is the charge separation. From triangle ABC we have: cos

cos 1 cosAlso: 4 4

where the electric dipole moment.o o

r d d

r r d

q d pr r r Vr r

p qd

θ

θ θπε πε

− +

− +

− ≈

≈ → ≈ =

= =

?

(24 - 6)

2

1 cos4 o

pVr

θπε

=

dqO A

2 2

Potential created by a line of charge oflength L and uniform linear charge densityλ at point P. Consider the charge element

at point A, a distance from O.From triangle OAP we have:

dq dx x

r d x

λ=

= + Here is the distance OPThe potential created by at P is:

ddV dq

(24 - 8)

Example :

dqO A

(24 - 8)

( )( )( )

2

2 2

2 20

2 2

0

2

2

2

2

2

1 14 4

4

ln

l

4

n ln4

n

l

o o

L

o

L

o

o

dx x d

dq dxdVr d xdxVd x

V x d x

V x

d

L

x

L

x

d

λπε πε

λπε

λπελπε

= =+

=+

⎡ ⎤= + +⎢ ⎥⎣ ⎦

⎡ ⎤= +

= + +

+ −⎢ ⎥⎣ ⎦

+∫

∫

:V from E gDeterminin•

sdρ

⊕q

Eρ

dVsdEqsdFdW ρρρρ

⋅=⋅=

dVsdEqdW

−=⋅=ρρ

) Calculus. of principle lFundamenta

(

path toparallel field ,||

||

||

∫−=−⇔

−=

dsEVV

dsdVEdsEif

if

kEjEiEE zyxˆˆˆ

general,In

++=ρ

,xVEx ∂∂

−= ,yVEy ∂∂

−= ,zVEz ∂∂

−= .)ˆˆˆ()ˆˆˆ( Vz

ky

jx

ikzVj

yVi

xVE

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

−=ρ

,ˆˆˆ)(

kz

jy

ixdel ∂

∂+

∂∂

+∂∂

≡∇ .VE −∇=ρ

EX. Check field for single charge

rqkV 0=

r

q

20

|| rqk

drdV

dsdVE =−=−=

rrqk

E ˆ2

0=ρ

EX. Check filed for uniformly charged thin ringknown

Q

V 0=∂∂

−=xVEx 0=

∂∂

−=yVEy

....=−=∂∂

−=dzdV

zVEz

( )F +

r( )F −

r

2

2 2

Many molecules such as H O have a permanent electric dipole moment. These are known as "polar" molecules. Others, such as O , N , etc the electric dipole moment is zero. These are known as "nonpolar" molecules. One such molecule is shownin fig.a. The electric dipole moment is zero because the center of the positive charge coincides with the center of the negative charge. In fig.b we show

what

pr

happens when an electric field is applied on a nonpolar molecule. The electric forces on the positive and nagative charges are equal in magnitude but opposite in direction.

Er

(24 - 9)Induced dipole moment

Work and Electric PotentialAssume a charge moves in an electric field without any change in its kinetic energyThe work performed on the charge isW = ΔV = q ΔV

U with Multiple Charges, finalIf there are more than two charges, then find U for each pair of charges and add themFor three charges:

The result is independent of the order of the charges

1 3 2 31 2

12 13 23e

q q q qq qU kr r r

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

Finding E From VAssume, to start, that E has only an xcomponent

Similar statements would apply to the y and zcomponentsEquipotential surfaces must always be perpendicular to the electric field lines passing through them

xdVEdx

= −

E and V for an Infinite Sheet of ChargeThe equipotential lines are the dashed blue linesThe electric field lines are the brown linesThe equipotential lines are everywhere perpendicular to the field lines

E and V for a Point ChargeThe equipotential lines are the dashed blue linesThe electric field lines are the brown linesThe equipotential lines are everywhere perpendicular to the field lines

E and V for a DipoleThe equipotential lines are the dashed blue linesThe electric field lines are the brown linesThe equipotential lines are everywhere perpendicular to the field lines

Electric Field from Potential, General

In general, the electric potential is a function of all three dimensionsGiven V (x, y, z) you can find Ex, Ey and Ez as partial derivatives

x y zV V VE E Ex y z

∂ ∂ ∂= − = − = −

∂ ∂ ∂

Electric Potential for a Continuous Charge Distribution

Consider a small charge element dq

Treat it as a point charge

The potential at some point due to this charge element is

edqdV kr

=

V for a Continuous Charge Distribution, cont.

To find the total potential, you need to integrate to include the contributions from all the elements

This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions

edqV kr

= ∫

V for a Uniformly Charged RingP is located on the perpendicular central axis of the uniformly charged ring

The ring has a radius a and a total charge Q

2 2e

ek QdqV k

r x a= =

+∫

V for a Uniformly Charged Disk

The ring has a radius a and surface charge density of σ

( )1

2 2 22 eV πk σ x a x⎡ ⎤= + −⎢ ⎥⎣ ⎦

V for a Finite Line of ChargeA rod of line ℓ has a total charge of Qand a linear charge density of λ

2 2

lnek Q aVa

⎛ ⎞+ += ⎜ ⎟⎜ ⎟

⎝ ⎠

l l

l

V for a Uniformly Charged Sphere

A solid sphere of radius R and total charge QFor r > R, For r < R,

eQV kr

=

( )2 232

eD C

k QV V R rR

− = −

Irregularly Shaped Objects, cont.

The field lines are still perpendicular to the conducting surface at all pointsThe equipotentialsurfaces are perpendicular to the field lines everywhere

Cavity in a ConductorAssume an irregularly shaped cavity is inside a conductorAssume no charges are inside the cavityThe electric field inside the conductor must be zero

Cavity in a Conductor, contThe electric field inside does not depend on the charge distribution on the outside surface of the conductorFor all paths between A and B,

A cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity

0B

B A AV V d− = − ⋅ =∫ E s

Corona DischargeIf the electric field near a conductor is sufficiently strong, electrons resulting from random ionizations of air molecules near the conductor accelerate away from their parent moleculesThese electrons can ionize additional molecules near the conductor

Corona Discharge, cont.This creates more free electronsThe corona discharge is the glow that results from the recombination of these free electrons with the ionized air moleculesThe ionization and corona discharge are most likely to occur near very sharp points

Millikan Oil-Drop Experiment –Experimental Set-Up

Millikan Oil-Drop ExperimentRobert Millikan measured e, the magnitude of the elementary charge on the electronHe also demonstrated the quantized nature of this chargeOil droplets pass through a small hole and are illuminated by a light

Active Figure 25.27

(SLIDESHOW MODE ONLY)

Oil-Drop Experiment, 2With no electric field between the plates, the gravitational force and the drag force (viscous) act on the electronThe drop reaches terminal velocity with FD = mg

Oil-Drop Experiment, 3When an electric field is set up between the plates

The upper plate has a higher potential

The drop reaches a new terminal velocity when the electrical force equals the sum of the drag force and gravity

Oil-Drop Experiment, finalThe drop can be raised and allowed to fall numerous times by turning the electric field on and offAfter many experiments, Millikandetermined:

q = ne where n = 1, 2, 3, …e = 1.60 x 10-19 C

Van de GraaffGenerator

Charge is delivered continuously to a high-potential electrode by means of a moving belt of insulating materialThe high-voltage electrode is a hollow metal dome mounted on an insulated columnLarge potentials can be developed by repeated trips of the beltProtons accelerated through such large potentials receive enough energy to initiate nuclear reactions

Electrostatic PrecipitatorAn application of electrical discharge in gases is the electrostatic precipitatorIt removes particulate matter from combustible gasesThe air to be cleaned enters the duct and moves near the wireAs the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles become chargedMost of the dirt particles are negatively charged and are drawn to the walls by the electric field

( )F +

r( )F −

r

2

2 2

Many molecules such as H O have a permanent electric dipole moment. These are known as "polar" molecules. Others, such as O , N , etc the electric dipole moment is zero. These are known as "nonpolar" molecules. One such molecule is shownin fig.a. The electric dipole moment is zero because the center of the positive charge coincides with the center of the negative charge. In fig.b we show

what

pr

happens when an electric field is applied on a nonpolar molecule. The electric forces on the positive and nagative charges are equal in magnitude but opposite in direction.

Er

(24 - 9)Induced dipole moment

V+V-

qCV

=

(25 - 2)

CapacitanceA system of two isolated conductors separated by an insulator (this can be vacuum or air) one with a charge +q and the other –q is known as a “capacitor”. The symbol used to indicate a capacitor is two parallel lines. We refer to the conductors as “plates”. We refer to the “charge” of the capacitor as the absolute value of the charge on either plate

Makeup of a CapacitorA capacitor consists of two conductors

These conductors are called platesWhen the conductor is charged, the plates carry charges of equal magnitude and opposite directions

A potential difference exists between the plates due to the charge

More About CapacitanceCapacitance will always be a positive quantityThe capacitance of a given capacitor is constantThe capacitance is a measure of the capacitor’s ability to store chargeThe farad is a large unit, typically you will see microfarads (μF) and picofarads (pF)

Definition of Capacitance

The capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the potential difference between the conductors

The SI unit of capacitance is the farad (F)

QCV

=Δ

Parallel Plate CapacitorEach plate is connected to a terminal of the batteryIf the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires

Parallel Plate Capacitor, contThis field applies a force on electrons in the wire just outside of the platesThe force causes the electrons to move onto the negative plateThis continues until equilibrium is achieved

The plate, the wire and the terminal are all at the same potential

At this point, there is no field present in the wire and the movement of the electrons ceases

Parallel Plate Capacitor, finalThe plate is now negatively chargedA similar process occurs at the other plate, electrons moving away from the plate and leaving it positively chargedIn its final configuration, the potential difference across the capacitor plates is the same as that between the terminals of the battery

Capacitance – Isolated Sphere

Assume a spherical charged conductorAssume V = 0 at infinity

Note, this is independent of the charge and the potential difference

4/ o

e e

Q Q RC πε RV k Q R k

= = = =Δ

Q−q

+

_⊕

Q+q

q

appqdW Vdq dqC

= =

2

0

222

21

21

21 CV

CVCUPE

CQdq

cqW

Q

EEapp ======⇒ ∫

2o 11 , fixed , .

2eA Q Qd A C Ud V C

ε↓↑ ↑

↑ = = = ↑：

2o 12 , fixed , .

2eA Q QA d C Ud V C

ε ↑↑

↓↑ = = = ↓：

d ε

A

dAC ε=

Energy stored in capacitor

1、Without battery：(Q fixed)