Electrical Charge

College Physics----by Dr.H.Huang, Department of Applied Physics 1

Hong Kong Polytechnic University Electrical Charge

Some Concepts:Electrical charge ( 电荷 ); Electric field ( 电场 ); Electric current ( 电流 ); Conductor (导体 ); Semiconductor (半导体 ); Insulator (绝缘体 ); Superconductor (超导体 );

Two Fundamental Properties of Electrical Charge:Charge is quantized: ; Charge is conserved:

Electrostatic Force:

permittivity constant (介电常数 ):

Note that this satisfies Newton's third law because it implies that exactly the same magnitude of force acts on q2 . Coulomb's law is a vector equation and includes the fact that the force acts along the line joining the charges. Like charges repel and unlike charges attract. Coulomb's law describes a force of infinite range which obeys the inverse square law, and is of the same form as the gravity force.

Shell Theorem:A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at its center. A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside the shell.

C1060.1 19e

22120 m/NC1085.8 229 /CmN1099.8 k



Spherical Conductors:If excess charge is placed on a spherical shell that is made of conducting material, the excess charge spreads uniformly over the external surface.

Example: The figure shows two particles fixed in a place: a particle of charge q1=+8q at the origin of an x axis and a particle of charge q2=-2q at x=L. At what point (other than infinitely far away) can a proton be placed so that it is in equilibrium? Is that equilibrium stable or unstable? Solution:

Lx

Lx

qq

x

qq pp 22

4

18

4

12

02

0

30

20

20 2

2

4

1

2

8

4

1

2If

L

xqq

xL

qq

xL

qqF

xLx

ppp

Example: The arrangement of six fixed charged particles, where =30, is shown. All six particles have the same magnitude of charge q; their electrical signs are as indicated. What is the net electrostatic force acting on q1 due to the other charges? Solution: 0



Example: Two identical, electrically isolated conducting spheres A and B are separated by a (center-to-center) distance a that is large compared to the spheres. Sphere A has a positive charge of +Q; sphere B is electrically neutral; and initially, there is no electrostatic force between the spheres. (a) Suppose the spheres are connected for a moment by a conducting wire. The wire is thin enough so that any net charge on it is negligible. What is the electrostatic force between the spheres after the wire is removed? (b) Next, suppose sphere A is grounded momentarily, and then the ground connection is removed. What now is the electrostatic force between the spheres?

Solution:

002:)(

16

1

4

12:)(

21

2

02

21

021

FqQqb

a

Q

a

qqFQqqa



Homework:1. The figure shows two charges, q1 and q2, held in a

fixed distance d apart. (a)What is the magnitude of the electrostatic force that acts on q1? Assume that q1=q2=20.0C and d=1.50m. (b)A third charge q3=20.0 C is brought in and placed as shown in the figure. What now is the magnitude of the electrostatic force on q1?

2. In the basic CsCl (cesium chloride) crystal structure, Cs+ ions form the corners of a cube and a Cl ion is at the cube’s center. The edge length of the cube is 0.40 nm. The Cs+ ions are each deficient by one electron (and thus each has a charge of +e), and the Cl ion has one excess electron (and thus has a charge of –e). (a) What is the magnitude of the net electrostatic force exerted on the Cl ion by the eight Cs+ ions at the corners of the cube? (b)If one of the Cs+ ions is missing, the crystal is said to have a defect; what is the magnitude of the net electrostatic force exerted on the Cl ion by the seven remaining Cs+ ions?


Hong Kong Polytechnic University Electric Fields

Electric FieldElectric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

Conversely, the electrostatic force acting on a particle with a charge q is

Electric Field of Point Charge:

0q

FE

204

1

r

QE source

n

ii

100 FF

n

ii

100 EE

EF q

A positive number is taken to be an outward field; the field of a negative charge is toward it. The electric field from any number of point charges can be obtained from a vector sum of the individual fields.



Example: The figure shows three particles with charges q1=+2Q, q2=2Q, and q3=4Q, each a distance d from the origin. What net electric field E is produced at the origin? Solution:

20

2220

321

330cos

422

4

1

30cos30cos30cos

d

Q

d

Q

d

Q

d

Q

EEEEx

030sin30sin30sin 321 EEEEx iE2

0

3

d

Q

321 EEEE

Example: The nucleus of a uranium atom has a radius R of 6.8 fm. Assuming that the positive charge of the nucleus is distributed uniformly, determine the electric field at a point on the surface of the nucleus due to that charge.

Solution: The atomic number Z=92

N/C109.24

1

4

1 212

02

0

R

Ze

R

QE



Electric Field Lines (电力线 ):Electric field lines provide a means for visualizing the direction and magnitude of electric fields. The electric field vector at any point is tangent to a field line through that point. The density of field lines in any region is proportional to the magnitude of the electric field in that region. Field lines originate on positive charges and terminate on negative charges. A few examples are shown below:

Electric Dipole (电偶极子 ) Field:

An electric dipole consists of two particles with charges of equal magnitude q but opposite sign, separated by a small distance d. It can be proved that the electric field generated by the dipole along the dipole axis is

where r is the distance to the center of the dipole r>>d.

30

//

2

4

1

r

pE



Dipole Moment (偶极矩 ):

p=qd is called the electric dipole moment. It is a vector with a direction pointing from the negative to the positive charge of the dipole.

The electric field in a plane perpendicular to the dipole axis that cuts through the dipole by half is

Dipole in an Electric Field:The electric field exerts a torque on the dipole

The dipole has a potential energy

where we choose the potential energy to be zero when the angle is 90.

Epτ

EpU

304

1

r

pE

sinsin

sinsin2

sin2

pEq

pqE

Fdd

Fd

F


Hong Kong Polytechnic University

Electric Field Due to a Charged Ring: (Optional)As shown in the figure, is the charge per unit length. The total charge on the ring is q.

Electric Field Due to a Charged Disk: (Optional)

For a uniformly charged disk with a radius R, we can cut the disk into small pieces of rings. Suppose the charge per unit area is , then for each piece of ring, it carries a charge of

Each ring contributes an electric field

The total field is

Electric Fields

23220

30

30

30

30

20

442

44

4

1cos

4

1cos

Rz

qz

r

qzR

r

zds

r

z

dsr

z

r

dqdEE

rdrdAdq 2 2322

02322

0

2

44

2

rz

rdrz

rz

rdrzdE

220

0

2322

0

124 Rz

zdrrz

zdEE

R



Homework:1. A clock face has negative point charges –q, 2q, 3q, …, 12q fixed at the positions of

the corresponding numerals. The clock hands do not perturb the net field due to the point charge. At what time does the hour hand point in the same direction as the electric field vector at the center of the dial? (Hint: Use symmetry).

2. In the figure a uniform, upward-pointing electric field E of magnitude 2.00103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L=10.0 cm and separation d=2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle =45.0 with the lower plate and has a magnitude of 6.00106 m/s. (a) Will the electron strike one of the plate? (b) If so, which plate and how far horizontally from the left edge?


Hong Kong Polytechnic University Electric Potential

Electrical Potential Energy:The electrostatic force is a conservative force. Thus we can define a change of the electric potential energy U of a charged particle in an electric field to be,

where W is the work done by the electrostatic force and is path independent.

If the potential energy is defined to be zero at infinity, the electric potential energy U of a point charge is U=W , where W is the work done by the electric field on the

point charge as the charge moves from infinity to the particular point.

Concept:In the figure, a proton moves from point i to point f in a uniform electric field directed as shown. (a) Does the electric field do positive or negative work on the proton? (b) Does the electric potential energy of the proton increase or decrease?

Electrical Potential:The electrical potential energy per unit charge is defined as the electric potential difference (电势差 ):

The electrical potential at a point is ; The unit is volt=joule per coulomb.

WUUU if

+f i

E

q

W

q

UVVV if

q

WV

VqU



Equipotential Surfaces (等势面 ):Adjacent points that have the same electric potential form an equipotential surface, which can be either an imaginary surface or a real, physical one.(i) No work is done by the electric field when a charge is moved on that surface.(ii) The electric field E is always directed perpendicularly to the equipotential surface.

Potential Calculated from the Field:

In a uniform electric field E, the potential difference is,

f

iif dqWqWVV sE

f

iif dVV sE f

idV sE

EdEdsVVf

iif EddsEVV

f

ccf 45cos



Potential due to a Point Charge:

Example: (a) What is the electric potential V at a distance r=2.1210-10 m from the nucleus of a hydrogen atom (the nucleus consists of a single proton)? (b) What is the electric potential energy U in electron-volts of an electron at the given distance from the nucleus? (c) If the electron moves closer to the proton, does the electric potential energy increase or decrease? Solution:

(c): V increases and U decreases.

rr

r

r

qdrddV

204

rErEr

qV

04

1

n

i i

in

ii r

qVV

101 4

1

r

dV

r

dqV

00 4

1

4

1

V78.64

1:)(

0

r

eVa

eV78.6:)( eVqVUb

Example: What is the potential at point P, located at the center of the square of point charges shown in the figure? Assume that d=1.3 m and that the charges are q1=+12 nC, q2=24 nC, q3=+31 nC, q4=+17 nC.

Solution: V350

24

1 4321

0

4

1

d

qqqqVV

ii



Example: (a) In figure a, 12 electrons are equally spaced and fixed around a circle of radius R. Relative to V=0 at infinity, what are the electric potential and electric field at the center C of the circle due to these electrons? (b) If the electrons are moved along the circle until they are nonuniformly spaced over a 120 are (figure b), what then is the potential at C? Solution:

012

4

1:)(

0

ER

eVa

R

eVb

12

4

1:)(

0

Potential of Line Charge: (Optional)

It can be found by superposing the point charge potentials of infinitesmal charge elements. It is an example of a continuous charge distribution.

22

22

022

0

ln44

1

daa

dbb

dx

dxV

b

a



Potential for Ring of Charge: (Optional)It can be found by superposing the point charge potentials of infinitesmal charge elements. The ring potential can then be used as a charge element to calculate the potential of a charged disc.

r

R

r

QV

00 24

1

Potential for Disc of Charge: (Optional)It can be found by superposing the point charge potentials of infinitesmal charge elements. The evaluation of the potential can be facilitated by summing the potentials of charged rings.

zRz

Rz

RdR

Rz

RdR

r

dqV

R

22

0

0 220

2200

2

2

2

4

1

4

1

Calculating E from V:

rErrrr d-dVVdVW

kjiEz

V

y

V

x

VzyxV

,,



Electric Potential Energy of a System of Point Charges:The electric potential energy of a system of fixed point charges is equal to the work that must be done by an external agent to assemble the system, bringing each charge in from an infinite distance.

For two point charges: For many point charges:

Example: Starting with the expression for the potential at any point on the axis of a charged disk, derive the expression for the electric field at any point on the axis of the disk. Solution:

Example: The figure shows three charges held in fixed positions by forces that are not shown. What is the electric potential energy of this system of charges? Assume that d=12 cm and that q1=+q, q2=4q, and q3=+2q, in which q=150 nC. Solution:

12

21

012 4

1

r

qqU

n

jii

n

j ij

ji

r

qqU

1 108

1

22

0

22

0

122 Rz

zzRz

dz

d

z

VEz

mJ174

1 133221

0

d

qq

d

qq

d

qqU



Homework:1. In a given lightning flash, the potential difference between a cloud and the ground is

1.0109 V and the quantity of charge transferred is 30 C. (a) What is the change in energy of that transferred charge? (b) If all the energy released by the transfer could be used to accelerate a 1000 kg automobile from rest, what would be the automobile’s final speed? (c) If the energy could be used to melt ice, how much ice would it melt at 0C? The heat of fusion of ice is 3.33105 J/kg.

2. For the charge configuration of the figure, show that V(r) for points such as P on the axis, assuming r>>d, is given by

(Note: the charges should be treated as point charges and their size being neglected.)

r

d

r

qV

21

4

1

0

+ +q +q +q P

d d

r


Hong Kong Polytechnic University Capacitance

Capacitor:A capacitor is a device that stores electric potential energy by storing separated positive and negative charges. It consists of two conductors separated by either vacuum or an insulating material. The simplest case is a parallel plate capacitor.

Capacitance is defined in terms of charge storage:

(Farad, F)

where, Q = magnitude of charge stored on each plate. V = voltage applied to the plates.

V

QC

A

QdEdV

A

QE

000

d

AC 0

Cylindrical Capacitor: (Optional)

The charge resides on the outer surface of the inner conductor and the inner wall of the outer conductor. Assume the length of the cylinder L>>b.

ab

LC

a

b

L

Qdr

rL

QEdrV

b

a

b

a

ln2

ln22

0

00



Spherical Capacitor: (Optional)

An Isolated Sphere: (Optional)

By taking the limits: aR and b,

Concept: For capacitors charged by the same battery, does the charge stored by the capacitor increase, decrease, or remain the same in each of the following situations? (a) The plate separation of a parallel capacitor is increased. (b) The radius of the inner cylinder of a cylindrical capacitor is increased. (c) The radius of the outer spherical shell of a spherical capacitor is increased.

ab

abC

ba

Qdr

r

QEdrV

b

a

b

a

0

02

0

411

44

RC 04

Example: The plates of a parallel-plate capacitor are separated by a distance d=1.0 mm. What must be the plate area if the capacitance is to be 1.0 F?

Solution:

Example: A storage capacitor on a random memory (RAM) chip has a capacitance of 55fF. If the capacitor is charged to 5.3 V, how many excess electrons are on its negative plate?

Solution:

28

0

m101.1 Cd

A

6108.1 e

CV

e

Qn



Capacitors in Parallel and Series:

Equivalent capacitance:

Example: (a) Find the equivalent capacitance of the combination as shown. Assume C1=12.0 F, C2=5.30 F, C3=4.50 F. (b) A potential difference V=12.5 V is applied to the input terminals. What is the charge on C1?

Solution: (a):

(b):

Example: A 3.55 F capacitor C1 is charged to a potential difference V0=6.30 V, using a 6.30 V battery. The battery is then removed and the capacitor is connected as in the figure to an uncharged 8.95 F capacitor C2. When switch S is closed, charge flows from C1 to C2 until the capacitors have the same potential difference V. What is the common potential difference? Solution:

(series)11

(parallel)11

n

i ieq

n

iieq CC

CC

312123

2112

111

CCC

CCC

F57.3

F3.17

123

12

C

C

V58.2C6.4412

123

12

1212123123

C

q

C

qVVCq

VCVCVC

qqq

2101

210

V79.121

10

CC

CVV



Potential Energy and Energy Density:The electric potential energy of a capacitor is the energy stored in the electric field between the two plates (electrodes). It is the work required to charge the capacitor.

The energy density is the potential energy per unit volume.

The above results hold generally for types of capacitors.

Example: An isolated conducting sphere whose radius R is 6.85 cm has a charge q=1.25 nC. (a) How much potential energy is stored in the electric field of this charged conductor? (b) What is the energy density at the surface of the sphere? (c) What is the radius R0 of an imaginary spherical surface such that half of the stored potential energy lies within it? Solution:

C

Qdq

C

qVdqdWW

QQ

2

2

00 2

2

2

1

2CV

C

QU

2

0

2

2

1

2

d

V

Ad

CV

Ad

Uu 2

02

1Eu

RRU

RR

qdrr

r

qdrruc

R

qEubn

R

q

C

qUa

R

R

R

R2

2

11

84

4

1

2

14:

J/m4.254

1

2

1

2

1:J103

82:

000

22

2

20

02

3

2

20

02

00

22

00



Dielectrics:Dielectric material contains polar molecules (or being polarized under an electric field), they will generally be in random orientations when no electric field is applied. An applied electric field will polarize the material by orienting the dipole moments of polar molecules. This decreases the effective electric field between the plates and will increase the capacitance of the parallel plate structure. The dielectric must be a good electric insulator so as to minimize any DC leakage current through a capacitor.

d

ACEEE peff

0

0

is called the dielectric constant of a material.

In a region completely filled by a material of dielectric constant , all electrostatic equation containing the permittivity constant 0 are to be replaced by 0.

The Guass’ law need to be generalized to,

qd AE0



Material

Air (1atm) 1.00054

Polystyrene 2.6

Paper 3.5

Transformer oil 4.5

Pyrex 4.7

Ruby mica 5.4

Porcelain 6.5

Silicon 12

Germanium 16

Ethanol 25

Water (20°C) 80.4

Water (25°C) 78.5

Titania ceramic 130

Strontium titanate 310

Room temperature dielectric constants for some materials

Example: A parallel-plate capacitor whose capacitance C is 13.5 pF is charged to a potential difference V=12.5 V between its plates. The charging battery is now disconnected and a porcelain slab is slipped between the plates. What is the potential energy of the device, both before and after the slab is introduced?

Solution:

pJ16022

pJ11002

1

22

2

i

ff

i

U

C

q

C

qU

CVU



Example: The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential difference V0 is applied between the plates. The battery is then disconnected, and a dielectric slab of thickness b and dielectric constant is placed between the plates as shown. Assume, A=115cm2, d=1.24cm, V0=85.5V, b=0.780cm, =2.61. (a) What is the capacitance C0 before the dielectric slab is inserted? (b) What free charge appears on the plates? (c) What is the electric field E0 in the gaps between the plate and the dielectric slab? (d) What is the electric field E1 in the dielectric slab? (e) What is the potential difference V between the plates after the slab has been introduced? (f) What is the capacitance with the slab in place?

Solution:

pF4.13:)(V3.52:)(

kV/m64.2:)(

kV/m90.6

:)(

pC720:)(

pF21.8:)(

10

01100

00

000

00

00

V

qCfbEbdEVe

A

qEqAEdd

A

qE

qAEdc

VCqbd

ACa

AE

AE


Hong Kong Polytechnic University

Homework (continued):1. The figure shows a variable “air gap” capacitor. Alternate

plates are connected together; one group is fixed in position and the other group is capable of rotation. Consider a pile of n plates of alternate polarity, each having an area A and separated from adjacent plates by a distance d. Show that this capacitor has a maximum capacitance of C=(n-1)0A/d.

2. In the figure, battery B supplies 12 V. (a) Find the charge on each capacitor first when only switch S1 is closed and (b) later when switch S2 is also closed. Take C1=1.0 F, C2=2.0 F, C3=3.0 F, and C4=4.0 F.

3. A parallel-plate capacitor of plate area A is filled with two dielectrics as shown in the figure. Show that the capacitance is

4. A parallel-plate capacitor of plate area A is filled with two dielectrics of the same thickness as shown in the figure. Show that the capacitance is

Capacitance

2210

d

AC

21

2102

d

AC

1 2

1

2


Hong Kong Polytechnic University Current and Resistance

Current:An electric current in a conductor is (ampere, A)

A current direction is the one in which positive charge carriers would move. The negative charges move in the opposite direction.

Current Density:

Drift of Charge Carriers:When an electric field E is established in a conductor, the charge carriers (assumed positive) acquire a drift speed vd in the direction of E,

where n is the number of charge carriers per unit volume.

Example: (a) The current density in a cylindrical wire of radius R=2.0 mm is uniform across a cross section of the wire and is given by J=2.0105 A/m2. What is the current through the outer portion of the wire between radial distance R/2 and R? (b) Suppose, instead, that the current density through a cross section varies with radial distance r as J=ar2, in which a=3.01011 A/m4 and r is in meters. What now is the current through the same outer portion of the wire?

Solution:

dt

dqi

AJ di

dnevJ

A1.72:)(

A9.14:)(

2

2

22

R

RdrrarJdAib

RRJJAia



Example: One end of an aluminum wire whose diameter is 2.5 mm is welded to one end of copper wire whose diameter is 1.8 mm. The composite wire carries a steady current i of 17 mA. (a) What is the current density in each wire? (b) What is the drift speed of the conduction electrons in the copper wire? Assume that, on the average, each copper atom contributes one conduction electron.

Solution:

Example: Consider a strip of silicon that has a rectangular cross section with width w=3.2 mm and height h=250 m, and through which there is a uniform current i of 5.2 mA. The silicon has a number of charge carriers (electrons) per unit volume n=1.51023 m-3. (a) What is the current density in the strip? (b) What is the drift speed? Solution:

Concept:The figure shows conduction electrons moving leftward through a wire. Are the following leftward or rightward: (a) the current i, (b) the current density J, (c) the electric field E in the wire?

m/s109.4:)(

A/m107.6A/m105.3:)(

7

2323

A

CuCud

CuCuAlAl

N

M

e

J

ne

Jvb

AiJAiJa

cm/s27:)(A/m6500:)( 2 ne

Jvb

wh

i

A

iJa d



Resistance: (ohm, )

Resistivity: unit of : m

Conductivity: (-1m-1)

The resistivity R of a conducting wire of length L and uniform cross-section A is

The resistivity for most materials changes with temperature

is the mean temperature coefficient of resistivity.

Ohm’s Law:The current through a device is always directly proportional to the potential difference applied to the device.

i

VR

JE

1

EJ

Material (m) (K-1)

Silver 1.6210-

8

4.110-3

Copper 1.6910-

8

4.310-3

Aluminium 2.7510-

8

4.410-3

Tungsten 5.2510-

8

4.510-3

Iron 9.6810-

8

6.510-3

Platinum 10.610-

8

3.910-3

Manganin 48.210-

8

-7010-3

Pure Silicon 2.5103

n-type Silicon 8.710-4

p-type Silicon 2.810-3

Glass 1010-1014

Fused quartz 1016

Resistivity of some materials at room temperature

A

LR

000 TT

A conducting material or device obeys Ohm’s law when the resistivity is independent of the magnitude and direction of the applied electric field.



Example: A rectangular block of iron has dimension 1.2 cm1.2 cm15 cm. (a) What is the resistance of the block measured between the two square ends? (b) What is the resistance between two opposite rectangular faces? Solution:

Power:

Example: A wire has a resistance R of 72 . At what rate is energy dissipated in each of the following situations? (1) A potential difference of 120 V is applied across the full length of the wire. (2) The wire is is cut in half, and a potential difference of 120 V is applied across the length of each half.

Solution:

Example: A wire of length L=2.35 m and diameter d=1.63 mm carries a current i of 1.24 A. The wire dissipates electrical energy at the rate P=48.5 mW. Of what is the wire made? Solution:

It’s aluminum.

A65.0:)(A100:)( A

LRb

A

LRa

RVRiiVP 22

W80022:)2(W200:)1( 22 RVPRVP

m1080.24

4 82

2

2

222

Li

Pd

d

Li

A

LiRiP



Semiconductor (optional):

Semiconductors are materials with intermediate conductivity between conductors and insulators.

From quantum mechanics, the electrons of an atom may occupy quantized energy levels. When a large amount of atoms form a solid, the discrete energy levels may be merged to form energy bands. The electrons are only allowed to sit within the energy bands but they cannot have an energy value within the gaps separating the gaps.

When excited (by thermal activation, for example), electrons can jump from the valence band to the conduction band and leave holes in the valence band. The conductivity of the material is enhanced.

Superconductors (optional):

Superconductivity is the lost of any conductivity at low temperatures. It was first discovered in mercury by K.Onnes in 1911. Since 1986, high temperature (~90K) superconductor in ceramics have been discovered and developed.



Homework:1. A charged belt, 50 cm wide, travels at 30 m/s between a source of charge and a sphere.

The belt carries charge into the sphere at a rate corresponding to 100 A. Compute the surface charge density on the belt.

2. In the figure, a resistance coil, wired to an external battery, is placed inside a thermally insulated cylinder fitted with a frictionless piston and containing an ideal gas. A current i=240 mA exists in the coil, which has a resistance R=550 . At what speed v must the piston, of mass m=12 kg, move upward to keep the temperature of the gas unchanged?

Electrical Charge

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