electrostaics coulomb force and electrical dipoles

8/9/2019 electrostaics coulomb force and electrical dipoles

1/31

(CharlesD.

Winters)

F IGURE 4.1 Rubbing a balloon against your hairon a dry day causes the balloon and your hair to

become electrically charged.

Electric Forces and Electric

Fields

Chapter 4


2/31

Rubber Rubber

(a)

F F

(b)

F

F

Rubber

+ + +

+

+

+

Glass+

F IGURE 4.2 (a) A negatively charged rubber rod, suspended by

an insulating thread, is attracted to a positively charged glass rod. (b) A

negatively charged rubber rod is repelled by another negatively charged

rubber rod.

F IGURE 4.3 When a glass rod is

rubbed with silk, electrons are

transferred from the glass to the silk.

Because of conservation of charge, each

electron adds negative charge to the silk,

and an equal positive charge is leftbehind on the rod. Also, because the

charges are transferred in discrete

bundles, the charges on the two objects

are e or 2e or 3e, and so on.


3/31


4/31

Suspensionhead

Fiber

B

A

F IGURE 4.6 Coulombstorsion balance, which was

used to establish the inverse-

square law for the electrostatic

force between two charges.


5/31

r

(a)F21

F12

q1

q2

(b)

F21

F12

q1

q2

r12

+

+

+

Figure 4.7

Two point charges separated

by a distance r exert a force

on each other given by

Coulombs law. Note that the

force exerted by q2 on

q1 is equal in magnitude andopposite in direction to the

force exerted byq1 on q2.

(a) When the charges are ofthe same sign, the force is

repulsive. (b) When the

charges are of opposite signs,

the force is attractive.

F:

12

F:

21


6/31

2.00 m

x

q1

x

q3q2 F13F23

2.00 x

+

+

F IGURE 4.8 (Example 4.1) Three

point charges are placed along the x

axis. If the net force on q3 is zero, the

force exerted byq1 on q3 must be

equal in magnitude and opposite in

direction to the force exerted byq2 on q3.

F

:

23

F:

13


7/31

(a) (b)

q0

q 0

>> q0

++

F IGURE 4.9 (a) For a small enough test

charge q0, the charge distribution on the

sphere is undisturbed. (b) If the test charge q0

were larger, the charge distribution on thesphere would be disturbed as a result of the

proximity ofq0.


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(b)

E

q

P

r

P

(a)

Fe

q

q0

rP

r

(c)

Feq

q0

P

r

(d)

E

q

r+

+

Figure 4.10

A test charge q0 at pointPis a distance rfrom a

point charge q. (a) Ifq is positive, the force on

the test charge is directed away from q. (b) For

the positive source charge, the electric field at

P points radially outward from q. (c) If q is

negative, the force on the test charge is

directed toward q. (d) For the negative source

charge, the electric field at P points radially

inward toward q.


9/31

P E

y

E1

E2y

r

aq

aq

x+

F I GURE 4.11 (Example4.3) The total electric field

at P due to two equal and

opposite charges (an electric

dipole) equals the vector sum

. The field is dueto the positive charge q, and

is the field due to the

negative charge q.

E:

2

E

:

1E

:

1 E

:

2

E:


10/31

r

qr

P

E

F IGURE 4.12 The

electric field at P due to a

continuous charge distributionis the vector sum of the fields

due to all the elements qof the charge distribution.

E:

E:


11/31

F IGURE 4.13 (Example 4.4) The electric field atPdue to a

uniformly charged rod lying along the xaxis. The field atPdue

to the segment of charge dq is kedq/x2. The total field atP isthe vector sum over all segments of the rod.

x

y

a

Px

dx

dq= dx

E


12/31

(a)

++

+

+

+

+

+

+

++

++

++

++

P dEx

dEdE

x x

r

q

a

(b)

++

+

+

+

+

+

+

+

+

++

+

++

+

dE2

dE1

2

F IGURE 4.14 (Example 4.5)

A uniformly charged ring of radius

a. (a) The field atPon the xaxis

due to an element of charge dq.

(b) The total electric field atP is

along the x axis. The perpendicu

lar component of the electric field

atPdue to segment 1 is canceled

by the perpendicular component

due to segment 2.

BA

F IGURE 4.15 Electric field lines

penetrating two surfaces. The

magnitude of the field is greater on

surface A than on surface B.


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(a)

q

(b)

q+

(DouglasC.

Johnson/CalPolyPomona)

F IGURE 4.16 The electric field lines for a point charge. (a) For a positive point charge, the lines are directed radially

outward. (b) For a negative point charge, the lines are directed radially inward. Note that the figures show only those field

lines that lie in the plane containing the charge. (c) The dark areas are small particles suspended in oil, which align with the

electric field produced by a small charged conductor at the center.

(c)

(a)

+ F IGURE 4.17 (a) The electric field lines for

two charges of equal magnitude and opposite sign

(an electric dipole). Note that the number of lines

leaving the positive charge equals the number

terminating at the negative charge. (b) Small

particles suspended in oil align with the electric

field.(DouglasC.


(b)


14/31

F I GURE 4.18 (a) The electric field lines

for two positive point charges. (The

locations A, B, and Care discussed in Quick

Quiz 4.5.) (b) Small particles suspended in

oil align with the electric field.(DouglasC.


(a)

+ +C

A

B

+2q q+

Figure 4.19

The electric field lines for a

point charge 2q and a

second point charge q.

Note that two lines leave the

charge 2qfor every one that

terminates on q.

(b)


15/31

+

+

+

+

+

+

E

vv= 0

q

x

+ +

F I GURE 4.20 (Example 4.6) A positive

point charge qin a uniform electric field

undergoes constant acceleration in thedirection of the field.

E:


16/31

(0, 0)

E

(x,y)

v

x

y

+ + + + + + + + + + + +

vii

Figure 4.21

An electron is projected horizontally into a uniform

electric field produced by two charged plates. The

electron undergoes a downward acceleration (opposite

), and its motion is parabolic while it is between the

plates.

E:


17/31

Electron

beam

plates

anode

gridgun

CA

plates

Fluorescentscreen

F IGURE 4.22 Schematic diagram of a cathode-ray tube. Electrons leaving the hot cathode C are accelerated to the anode A. In

addition to accelerating electrons, the electron gun is also used to focus the beam of electrons, and the plates deflect the beam.


18/31

A

in Equation 4.18 is then 90.

Area = A

E

F IGURE 4.23 Field lines of a

uniform electric field penetrating aplane of area A perpendicular to the

field. The electric flux E through this

area is equal toEA.

A

A = Acos

E

Normal

F IGURE 4.24 Field lines for a uniform electric

field through an area Awhose normal is at an angle

to the field. Because the number of lines that go

through the shaded area A is the same as thenumber that go through A, we conclude that the

total flux through A is equal to the flux through A

and is given by E EAcos .

Ai

E i

i

F IGURE 4.25 Asmall element

of a surface of area Ai. The

electric field makes an angle iwiththe normal to the surface (the

direction of ), and the flux

through the element is equal to Ei

Aicos i.

A:

i


19/31

Figure 4.26

A closed surface in an electric field. The area

vectors are, by convention, normal to

the surface and point outward. The flux

through an area element can be positive

(element 1), zero (element 2), or negative

(element 3).

A:

i

A1

A3

A2

E

En

En

EE


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y

z

x

E

dA2

dA1

dA3

dA4

F IGURE 4.27 (Example 4.8) A

hypothetical surface in the shape of a cube in

a uniform electric field parallel to the x axis.

The net flux through the surface is zero. Sideis the bottom of the cube and side is

opposite side .

Spherical

surface

r

q

A

E

i+

F IGURE 4.28 A spherical surface

of radius r surrounding a point

charge q. When the charge is at the

center of the sphere, the electric fieldis normal to the surface and constant

in magnitude everywhere on the

surface.


21/31

S3

S2

S1

q

F IGURE 4.29 Closed surfaces of

various shapes surrounding a charge q.

The net electric flux through eachsurface is the same.

F IGURE 4.30 Apoint charge located

outside a closed surface. The number of

lines entering the surface equals the

number leaving the surface.

q


22/31

S

q1

q2

q3 S

S

q4

Figure 4.31

The net electric flux through any closed

surface depends only on the charge

inside that surface. The net flux through

surface S is ql/e0, the net flux throughsurface S9 is (q2 1 q3)/e0, and the net

flux through surface S0 is zero. Charge

q4 does not contribute to the flux

through any surface because it is outsideall surfaces.


23/31

auss ansurface

r

q

dA

E

+

F IGURE 4.32 (Example 4.9)

The point charge q is at the

center of the spherical gaussian

surface, and is parallel to at

every point on the surface.

dA:

E:


24/31

(a)

Gaussiansphere

(b)

Gaussianspherer

a

r

a

F IGURE 4.33(Interactive Example 4.10) Auniformly charged insulating sphere of radius aand total

charge Q. (a) For points outside the sphere, a large,

spherical gaussian surface is drawn concentric with the

sphere. In diagrams such as this one, the dotted line

represents the intersection of the gaussian surface withthe plane of the page. (b) For points inside the sphere, a

spherical gaussian surface smaller than the sphere is

drawn.


25/31

Gaussian

surface

+

++

+++

E

dA

r

(a)

E

(b)

F IGURE 4.35 (Example 4.11) (a) An infinite line of

charge surrounded by a cylindrical gaussian surface

concentric with the line charge. (b) An end view shows

that the electric field on the cylindrical surface is

constant in magnitude and perpendicular to the surface.

a

E

ar

E=keQ

r2

E=keQ

a3r

F IGURE 4.34(Example 4.10) A plot ofEversus rfora uniformly charged insulating sphere. The electric field

inside the sphere (r a) varies linearly with r. The

electric field outside the sphere (r a) is the same as

that of a point charge Qlocated atr 0.


26/31

E

++

++

++

+

++

++

++

++

+

++

++

+

++

+

++

++

+

++

++

++

+

A

Gaussiansurface

E

F IGURE 4.36 (Example 4.12) A

cylindrical gaussian surface penetrating an

infinite sheet of charge. The flux is EA

through each end of the gaussian surface and

zero through its curved surface.


27/31

+

+

++

+

++

+

E E

F I GURE 4.37 Aconducting

slab in an external electric field

. The charges induced on the

surfaces of the slab produce an

electric field that opposes theexternal field, giving a resultant

field of zero inside the

conductor.

E:

Gaussiansurface

F I GURE 4.38 An isolated

conductor of arbitrary shape.

The broken line represents a

gaussian surface just inside the

physical surface of the

conductor.


28/31

A

++ + +

+

+

+

+

+++++++ +

++

+

+

E

F IGURE 4.39 A gaussiansurface in the shape of a small

cylinder is used to calculate the

electric field just outside a

charged conductor. The flux

through the gaussian surface isEA.


29/31

0.500 m

7.00C

2.00C 4.00C

60.0

y

+

+

Figure P4.5


30/31

d

+3q +q

Figure P4.8

0.100 m

x

5.00 nC

0.300 m

6.00 nC

y

aa

a

a

q

3q 4q

2q

Figure P4.15

2a

x

q q

Figure P4.16

1.00 m

2.50C 6.00C

Figure P4.11

Figure P4.13


31/31

y

y

dx

x

P

O

0

Figure P4.21

q2

q1

Figure P4.23

a

q

a a

P +

++

electrostaics coulomb force and electrical dipoles

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