CHAPTER 23 : ELECTRIC FIELDS

23.1 : PROPERTIES OF ELECTRIC CHARGES

•Existence of electric forces and charges :•Running a comb through your hair on a dry day•Glass or rubber - rubbed with silk or fur•Inflated balloon rubbed with wool – adheres to a wall•Rubbing shoes on a wool rug

• Electrified / Electrically charged• Electric charges - positive

- negative• Like charges – repel (+ve with +ve / -ve with –ve) Unlike charges – attract (+ve with –ve)

• Electric charge is always conserved• Electric charge is quantized

Figure (23.1)

Rubber

F F

Glass

(a) (b)

F

F

Rubber

Rubber

A negatively charged rubber rod suspended by a thread is attracted to a positively charged glass rod

A negatively charged rubber rod is repelled by another negatively charged rubber rod.

FIGURE (23.1)

Quiz (23.1) : If you rub an inflated balloon against your hair, the two materials attract each other (Fig. 23.2). Is the amount of charge present in the balloon and your hair after rubbing (a) less than, (b) the same as, or (c) more than the amount of charge present before rubbing?

23.2 : INSULATORS AND CONDUCTORS

• Electrical conductors – materials in which electric charges move freely.• Electrical insulators – materials in which electric charges cannot move freely.• Conductor – e.g.: copper, aluminum, silver• Insulator – e.g.: glass, rubber, wood• Semiconductors - their electrical properties are somewhere between those of

insulators and those of conductors.

- Silicon and germanium.

- Electrical properties of semiconductors can be changed.• Grounded – when a conductor is connected to the earth by means of a

conducting wire or pipe.• Induction – process of charging a conductor (Fig. 23.3).

•A process similar to induction in conductors takes place in insulator.

(a)

Charged object Induced charges

Insulator

FIGURE (23.4)

Quiz 23.2 : Object A is attracted to object B. If object B is known to be positively charged, what can we say about object A? (a) It is positively charged, (b) It is negatively charged, (c) It is electrically neutral, or (d) Not enough informatin to answer.

23.3 : COULOMB’S LAW

• Coulomb’s experiment showed the electric force between two stationary charge of particles :

i) ii) iii) Opposite sign - attractive Same sign - repulsive

• Coulomb’s Law

2r

1F 21qqF

oe 4

1k

Permittivity of free space, = 8.8542 x 10-12 C2/Nm2o

2

21ee r

qqkF

Coulomb constant, ke= 8.9875 x 109 Nm2/C2

(23.1)

Particle Charge (C) Mass (kg)

Electron (e) - 1.6021917 x 10-19 9.1095 x 10-31

Proton (p) + 1.6021917 x 10-19 1.67261 x 10-27

Neutron (n) 0 1.67492 x 10-27

Table 23.1

Example (23.1) : The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3 x 10-11 m. Find the magnitudes of the electric force and the gravitational force between the two particles.

e pr

5.3 x 10-11 m

•Coulomb’s Law – vector quantity

Quiz (23.2) : Object A has a charge of +2C, and object B has a charge of +6 C. Which statement is true?

(a)FAB = -3FBA

(b)FAB = -FBA

(c) 3FAB = -FBA

r̂r

qqkF

221

e12

1221 FF

(23.2)

Example (23.2) : Find the Resultant Force

Consider three point charges located at the corners of a right triangle (Fig. 23.7), where q1 = q3 = 5.0 C, q2 = -2.0 C, and a = 0.10 m. Find the resultant force exerted on q3.

y

x

-

+

+q2

q1

q3

a

a FIGURE (23.7)

4131211 FFFF

a2

23F 13F

Example (23.3) : Where is the Resultant Force Zero?

Three point charges lie along the x-axis as shown in Figure (23.8). The positive charge q1 = 15.0 C is at x = 2.00 m, the positive charge q2 = 6.00 C is at the origin, and the resultant force acting on q3 is zero. What is the x coordinate of q3?

2.00 m

2.00 - xx

+ - + x

q2 q3 q1

FIGURE (23.8)

23F 13F

Example (23.4) : Find the Charge on the Spheres

Two identical small charged spheres, each having a mass of 3.0 x 10-2 kg, hang in equilibrium as shown in Fig. (23.9a). The length of each string is 0.15 m, and the angle is 5.0o. Find the magnitude of the charge on each sphere.

(a)

L L

aq q

L = 0.15 m = 5.0o

T cos

T sin

(b)FIGURE (23.9)

T

eF

gm

23.4 : THE ELECTRIC FIELD

•The strength (magnitude) of the electric field, at a point in space is defined as the electric force Fe acting on a positive test charge, qo placed at that point divided by the magnitude of the test charge :

•Newtons per coulomb (N/C)

• is a vector - magnitude

- direction

• An electric field exists at a point if a test charge at rest at that point experiences an electric force.

•Charge qo is small enough that it does not disturb the charge distribution.

FIGURE (23.10)

Qqo

+

+

+

++

+++

+ + +

+

- ---- --

- - --

-

-

--

---

- --

-

--

- -

+ +qo

(a) (b)FIGURE 23.11

E

E

E

o'o qq

o

e

q

FE (23.3)

• To calculate electric field at a point P due to a group of point charges :

1) Calculate the E vectors at P individually using Equation (23.4).

2) Add them vectorially

At any point P, the total electric field due to a group of charges equals the vector sum of the electric fields of the individual charges.

r̂r

qqkF

2o

ee

o

e

q

FE

r̂r

qkE

2e (23.4)

E

ii

2i

ie r̂

r

qkE

(23.5)

Quiz 23.4 : A charge of +3 C is at a point P where the electric field is directed to the right and has a magnitude of 4 x 106 N/C. If the charge is replaced with a -3 C charge, what happens to the electric field at P?

Example (23.5) : Electric Field Due to Two Charges

A charge q1=7.0C is located at the origin, and a second charge q2=-5.0 C is located on the x-axis, 0.30 m from the origin (Fig. 23.13). Find the electric field at the point P, which has coordinates (0, 0.40)m.

y

+ -

1E

E

P

2E

0.40m 0.50m

0.30mx

q1 q2

FIGURE (23.13)

Exercise : Find the electric force exerted on a charge of 2.0x10-8C located at P.

Example (23.6) : Electric Field of a Dipole

An electric dipole is define as a positive charge q and a negative carge –q separated by some distance. For the dipole shown in Figure 23.14, find the electric field at P due to the charges, where P is a distance y >> a from the origin.

E

23.5 : ELECTRIC FIELD OF A CONTINUOUS CHARGE DISTRIBUTION

qr̂

r

PE

FIGURE (23.15)

Procedure :

1) Devide the charge distribution into small elements, each of which contains a small charge q (Fig. 23.15)

2) Use Eq. (23.4) to calculate the electric field due to one of these elements at a point P.

3) Evaluate the total field at P due to the charge distribution by summing the contributions of all the charge elements (applying the superposition principle).

r̂r

qkE

2e

ii

2i

ie r̂

r

qkE

r̂r

dqkr̂

r

qlimkE

2eii

2i

ie

iq 0

(23.6)

The volume charge density,

V

Q

Unit : C/m3

The surface charge density,

A

Q

Unit : C/m2

The linear charge density,

Q

Unit : C/m

dA

dQ

dV

dQ

d

dQ

Example (23.7) : The Electric Field Due to a Charged Rod

A rod of length has a uniform positive charge per unit length and a total charge Q. Calculate the electric field at a point P thant is located along the long axis of the rod and a distance a from one end (Fig. 23.16).

Example (23.8) : The Electric Field of a Uniform Ring of Charge

A ring of radius a carries a uniformly distributed positive total charge Q. Calculate the electric field due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring (Fig. 23.17a).

Exercise (Example (23.8)) : Show that at great distances from the ring (x>>a) the electric field along the axis shown in Figure (23.17) approaches that of a point charge of magnitude Q.

Example (23.9) : The Electric Field of a Uniformly Charged Disk

A disk of radius R has a uniform surface charge density . Calculate the electric field at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk (Fig. 23.18).

23.6 : ELECTRIC FIELD LINES

• Drawing lines that follow the same direction as the electric field vector at any point

• Electric field lines.

• Electric filed in any region of space :

1) The electric field vector is tangent to the electric field line at each point.

2) The magnitude of the electric field – depends on the number of lines per unit area through a surface perpendicular to the lines.

3) Nonuniform

E

Positive Point Charge Negative Point Charge

REPELLED positive test charge ATTRACT positive test charge

Outward Inward

+ -

(a) (b)

q -q

Rules for drawing electric field lines :

1) The lines must begin on a positive charge and terminate on a negative charge.

2) The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.

3) No two field lines can cross.

FIGURE (23.21(a))

FIGURE (23.22(a))

+

+

A

C

B

Quiz (23.5) : Rank the magnitude of the electric field at points A, B and C shown in Figure (23.22(a)) (greatest magnitude first).

+ -+2q

-q

FIGURE (23.23)

23.7) : MOTION OF CHARGED PARTICLES IN A UNIFORM ELECTRIC FIELD

•When a particle of charge q and mass m is placed in an electric field , the electric force exerted on the charge is q .

•The only force exerted on particle – it must be the net force and cause the particle to accelerate.

•Newton’s second law applied to the particle, gives :

E

E

amEqFe

•The acceleration of the particle is therefore

m

Eqa

(23.7)

• is uniform (constant in magnitude and direction) – acceleration is constant.

• Positive charge - its acceleration is in the direction of the electric field.

• Negative charge – its acceleration is in the direction opposite the electric field.

E

Example (23.10) : An Acceleration Positive Charge

A positive point charge q of mass m is released from rest in a uniform electric field directed along the x-axis, as shown in Figure (23.24). Describe its motion.

E

+

+

+

+

+

+

-

-

-

-

-

-

E

x

+ +q

0v v

FIGURE (23.24)

FIGURE (23.25)

+ + ++ + + +

- - - - - - - - -

+

E(0,0)

-ii

(x,y)-

y

x

•The electric field, is in the positive y direction – the acceleration of the electron is in the negative y direction.

E

jm

eEa

(23.8)

•Acceleration is constant – equations of kinematic in two dimensions (Chapter 4) with :

ixi 0yi •After the electron has been in the electric field for a time t, the components of its velocity are :

constantix (23.9)

tm

eEta yy (23.10)

•Its coordinates after a time t in the field are :

tx i

22y2

1 tm

eE

2

1tay

(23.12)

(23.11)

• Subtituting the value (Eq. 23.11) into Eq. (23.12), we see that y is

propotional to x2. i

xt

•Hence, the trajectory is a parabole.

•After the electron leaves the field, it continues to move in a straight line in the direction of in Figure (23.25), obeying Newton’s first law, with a speed .

i

Example (23.11) : An Accelerated Electron

An electron enters the region of a uniform electric field as shown in Figure (23.25), with = 3.00 x 106 m/s and E = 200 N/C. The horizontal length of the plates is = 0.100 m.

(a) Find the acceleration of the electron while it is in the electric field.

(b) Find the time it takes the electron to travel through the field.

(c) What is the vertical displacement y of the electron while it is in the field?

i

Exercise : Find the speed of the electron as it emerges from the field (3.22x106m/s).