PHY2049General Physics II

ELECTRICITY, MAGNETISM & LIGHT

• Chapter 21The electric field: I discrete charge distributions

• Chapter 22The electric field: II continuous charge distributions

• Chapter 23Electric potential

• Chapter 24Electrostatic energy and capacitance

• Chapter 25Electric current and direct-current circuits

• Chapter 26The magnetic field

• Chapter 27Source of the magnetic field

• Chapter 28Magnetic induction

• Chapter 29Alternating-current circuits

• Chapters 30/31Electromagnetic waves and the properties of light

• Chapter 32Optical images

CHAPTER 21

THE ELECTRIC FIELDI DISCRETE CHARGE DISTRIBUTIONS

• Electric charges

• Conductors and Insulators

• Coulomb’s Law

• Electric field

• Electric field lines

• Point charges in an electric field

• Electric dipoles

Rubber and wax rods rubbed with fur.

Glass rod rubbed with silk.

EFA02VD1.MOV

Rubber

Glass

! F ! F

Rubber

Rubber ! F

! F

Rubber rod rubbed with fur

REPELS

REPELS

ATTRACTS

Glass rod/silk ⇒ Glass rod ( +ve)Silk ( −ve)

Rubber rod/fur ⇒ Rubber rod ( −ve) Fur ( +ve)

Wax rod/fur ⇒ Wax rod ( −ve) Fur ( +ve)

Two types of charges: positive & negative(Benjamin Franklin (1706-1790)

Like charges ⇒ REPEL

Unlike charges ⇒ ATTRACT

Rubber rod rubbed with fur

Glass rod rubbed with silk

Rubber rod rubbed with fur

Wax rod rubbed with fur

Rubber rod rubbed with fur

Electrostatic series

GlassMicaWool

QuartzFurSilkFelt

CottonWoodCork

AmberRubberSulfur

Positive

When in contact, each material becomes more positivethan the ones below it.

Example: glass rubbed with silkglass ⇒ positivesilk ⇒ negative

∴ Total charge Q ⇒ newhere n is an integer and e is the smallest, indivisible charge - the charge of one electron.

Typical values of n ⇒ 1010 → 1016.

Charge is measured in Coulombs (later).

e = −1.60 ×10−19 C,

∴Q ~10−3 → 10−9C.

Electric charge is “quantized” ...

--- --

---

-- - Each individual

charge ⇒ e

DISCUSSION PROBLEM [21.1]

We are used to thinking that various quantities like time, speed, length, vary smoothly, i.e., continuously. Can you think of several examples of “quantized” quantities, i.e., those that do not change continuously but from some value N to N±1?

Normally, in an insulator, like a piece of paper, the molecules are orientated randomly. Furthermore, there are the same numbers of positive and negative charges in each molecule and so, overall, the object is electrically NEUTRAL.

+--+

Plastic has a greater affinity for electrons than fur. So, if a plastic rod is rubbed with fur, the rod becomes negatively charged while the fur becomes positively charged.

Silk has a greater affinity for electrons than glass. So, if a glass rod is rubbed with silk, the rod becomes positively charged while the silk becomes negatively charged.

- -- -++++ Plastic rod

Fur

electrons

+++ +----

Glass rod

Silk

electrons

If a charged rod is brought near to a piece of paper, the charges in the molecules shift slightly so that each molecule has a slightly more positive end - which in this case is

attracted towards the rod - and a slightly more negative end. However, the molecules do not move; they become polarized.

----

Charged rod

Attractiveforce

+- +- +- +-+-+-

+-+-

+-+-

+-

+

“normal”

----

--

--

--

--

--

- -

+

“polarized”

+

--

---

--

- - ----

- --

-

Note: dry air ⇒ insulator moist air ⇒ conductor

Therefore, conductors cannot be given a permanent charge by the same method as insulators as the charges can move away from each other. Conductors can be charged by another method ...

Introduce charges in an object.

If the object is an insulator, the charges remain localized.

If the object is a conductor, the charges repel each other and

rapidly disperse.

Comparison between insulators and conductors.

In an uncharged conductor, like a metal sphere, there are equal numbers of positive and negative charges and so there is no net charge: the object is electrically NEUTRAL.

Negative charges can move freely in a conductor so when a negatively charged rod is brought near, negative charges tend to move away from the rod and cause the object to become polarized. The positive charges (ions) remain fixed.

Note: charges are not created, they are REDISTRIBUTED

8 − 8 + net charge: 0

Charging a conductor:

A conductor can be given a permanent +ve charge by removing some of the −ve charges. The latter can be “carried” away through a wire attached to the ground or by touching the object with a finger ... called “grounding”.

Charging by induction

groundingsymbol

4 − 8 + net charge: 4 +

Wall outlet

Ground (or Earth)connectors

Grounding terminal(usually green)

Grounding rod

Ground

The ground connectors of a wall outlet are connected to a long metal spike (called a grounding rod) driven into the ground.

Conceptual question:

Question 21.1: Two, uncharged, insulated conducting spheres A and B are in contact. A positively charged rod is brought near A, but does not touch A. A and B are then separated while the charged rod remains near A. After the rod is removed, what are the charges on A and B?

+ A B

+++

Sphere A will be negatively charged and sphere B will be positively charged. This is another way of charging a conductor, instead of by induction.

B + +

+ + A

- -

- -

++++ B

+ + + +

A - - - -

++++ B

+ + + +

A - - - -

When two charged conducting spheres with the same radii but different charges touch, the charges redistribute so the charges are the same on each sphere. This is only true if the radii of the spheres are the same. In chapter 23 we deal with the case when the radii are different.

Q1 Q2

Q1 + Q2

2 Q1 + Q2

2

+ +

-

! F 21

q1 q2

! F 12

+ q1 q2

! F 21

! F 12

! F 21 = −

! F 12

q1

q2

r ! r

Coulomb’s Law.

Charles Augustin de Coulomb (1736-1806)

! F = k

q1 q2! r 2

! F = k

q1q2! r 2

ˆ r

• If q1 and q2 are of the same sign, ! F > 0

REPULSIVE• If q1 and q2 are of opposite sign,

! F < 0

ATTRACTIVE

F = k

q1q2r2

Charges q1 and q2 ⇒ Coulombs (C)Distance r ⇒ meters (m)

Force ⇒ Newtons (N)Then:

k = 8.98755×109 Nm2 C2 (≈ 9 ×109) = 1

4πε!,

where ε! = 8.854 ×10−12 C2 Nm2 , is called the

permittivity of free space (i.e., vacuum). What if the charges are in a medium?

... replace ε! ⇒ ε = κε!,where κ is called the dielectric constant.

κAir 1.00059Oil 2.2Methyl Alcohol 33.6Water 80.9

∴k ⇒ k κ , so the force is reduced. Normally, for air we take κ = 1.

Question 21.2: (a) In an atom of hydrogen the electron and proton are separated by an average distance of

5.3×10−11m. What is the magnitude of the mutual

electrostatic force of attraction between the two?

(b) Calculate the ratio of the electric force to the gravitational force between the two particles.

(a)Proton and an electron a distance 5.3 ×10−11mapart.

q1 = +1.6×10-19C and q2 = −1.6×10-19C

! F E = k

q1q2r2

= 9 ×109 ×

(+1.6 ×10−19)× (−1.6×10−19)(5.3 ×10−11)2

= 8.19 ×10−8 N (attractive).

(b) FG = G

memp

r2

mp = 1.67 ×10−27 kg me = 9.11×10−31kg

= 6.67 ×10−11 ×

(1.67 ×10−27)× (9.11×10−31)(5.3×10−11)2

= 3.61×10−47 N (always attractive)

∴

FEFG

=8.19 ×10−8

3.61×10−47 = 2.27 ×1039 !!

+ - 5.3 ×10−11m q1 q2

See ... the gravitational force is very weak compared with the electric force.

... then how come gravity is the dominant force between objects like planets and stars?

Ah well! It tells us that there must be an upper limit on the amount of charge on a star or planet.

M1,Q1 M2,Q2 R

Since FE << FG, then kQ1Q2 << GM1M2,

i.e., Q <<

Gk

M or Q << 8.6 ×10−11M.

For the Earth: M ≈ 6 ×1024 kg

∴Q << 5.2 ×1014C

(Actual value ... ~106C)

FE = k

Q1Q2R2

FG = GM1M2

R2

Question 21.3: The charges shown below are all point charges. Which configuration produces the largest repulsive force?

r

+q +2q +q +q

r2

A B C

D E

-2q +2q

r2

1.5r

+2q +2q

2r

+2q +2q

Answer B.

r

+q +2q

A

+q +q

r2

B

1.5r

+2q +2q

D

2r

+2q +2q

E

F = k

2q2

r2 = 2 kq2

r2⎛ ⎝ ⎜

⎞ ⎠ ⎟

Force is attractive, notrepulsive!

F = kq × qr2( )2 = 4 k

q2

r2⎛ ⎝ ⎜

⎞ ⎠ ⎟

F = k

2q × 2q(1.5r)2

= 1.78 kq2

r2⎛ ⎝ ⎜

⎞ ⎠ ⎟

F = k

2q × 2q(2r)2

= kq2

r2⎛ ⎝ ⎜

⎞ ⎠ ⎟

-2q +2q

C r2

When there are several electric forces acting we must take the VECTOR SUM, viz:

The resultant ! F R =

! F 23 +

! F 13.

In this example if 1 and 2 are fixed ... then 3 will

experience an initial acceleration in the direction of ! F R

given by Newton’s 2nd Law ! a =! F R

m3.

(Fixed)

(Fixed)

! F 23

! F 13

! F R

2

1

3

+

+ +

Question 21.4: Three point charges, each of magnitude

3.00nC, are at separate corners of a square whose sides are 5.00cm long. The two charges on opposite corners are positive and the third charge is negative. Find the force by these charges on a fourth point charge

q4 = +3.00nC at the remaining corner.

i

j

+3×10−9C

+3×10−9C

−3×10−9C

+ ! F 14

! F 24

! F 34

5 cm

5 cm

1

2 3

4

+

+

−

135"

! F 14 = k

(3×10−9)× (3×10−9)(5 ×10−2)2

ˆ i = 3.24 ×10−5N ˆ i .

! F 24 = −k

(3 ×10−9)× (3 ×10−9)(5×10−2)2

ˆ j = −3.24 ×10−5N j .

! F 34,x =

! F 34 cos135" ˆ i = −

! F 34 cos45" i .

! F 34,y =

! F 34 sin135" j =

! F 34 sin45" j .

We will need the distance between 3 and 4 :

= (5 ×10−2)2 + (5 ×10−2)2 = 7.07 ×10−2 m.

i

j

+3×10−9C

+3×10−9C

−3×10−9C

+ ! F 14

! F 24

! F 34

5 cm

5 cm

1

2 3

4

+

+

−

135"

∴! F 34,x = −k

(3 ×10−9) × (−3×10−9)(7.07 ×10−2 )2

× 0.707 ˆ i

= −1.14 ×10−5N ˆ i

and

! F 34,y = k

(3×10−9) × (−3×10−9)(7.07 ×10−2)2 × 0.707 j

= 1.14 ×10−5N j

∴! F R =

! F 14 +

! F 24 + (! F 34,x +

! F 34,y)

= 3.24 ×10−5ˆ i − 3.24 ×10−5ˆ j

+(−1.14 ×10−5ˆ i +1.14 ×10−5ˆ j )

= (2.1×10−5ˆ i − 2.1×10−5ˆ j ) N

The electric force between these two charges is:

! F = k

Q1Q2! r 2

ˆ r .

We say it is the electric field produced by Q1 that Q2

“feels” and produces the force on Q2 (and vice versa).

The electric field of Q1 at ! r (i.e., at the position of Q2) is

defined as:

! E (! r ) = Force on Q2

Q2=! F

Q2= k

Q1! r 2

ˆ r

Note also, the force on Q2 is ! F = Q2

! E .

So, we can find the electric field at ! r , ! E (! r ), due to Q1 by

using a “test charge” q" (of known charge) in place of Q2

and measuring ! F !

The Electric Field:

Q2 Q1 ! r +++

+

+

+ +

++

The gravitational force that the apple experiences due to the presence of the Earth is its weight:

! F = m! g .

We say it is the gravitational field produced by M (the Earth) that m (the apple) “feels” and produces the force on the apple. Using the same definition we used for the electric field, the gravitational field of M at r (i.e., the position of m) is:

Force on m

m=

m! g m

= ! g

so, ! g is “equivalent” to

! E .

Note also that: ! F G = m! g and

! F E = q

! E .

However, ! F G is always attractive whereas

! F E can be

attractive OR repulsive.

cf: the Gravitational Field:

m M

r

Earth

apple

So, we can determine both the magnitude and direction of

! E by using a “test charge” q" to find

! F , since

! E =! F q"

However ... we must assume that our test charge does not affect the charge Q, i.e.,

(a) with no test charge, the charge is distributed uniformly over the sphere.

(b) if q" is large, it may affect the way the charge Q is distributed.

Thus, the physically correct definition of the electric field is:

! E = Limit q"→0

! F

q"⎛ ⎝ ⎜ ⎞

⎠ ⎟

But we do not need to be this specific ... we always assume that a test charge q" is very small. Furthermore, we always assume that the test charge q" is positive, so

that ! E ℓ

! F .

+

+

+ ++ +

+ +

+

+++

++

+ +

+ q"

Electric field lines

An electric field can be shown as a set of lines that are parallel to the electric field vector at any point in space. These lines have the following properties:

• The electric field vector ! E is tangent to the electric

field line at every point.• Electric field lines are directed away from a +vecharge and towards a −ve charge.• The number of lines per unit area perpendicular to the lines is proportional to the magnitude of the electric field in that region.• Electric field lines cannot cross.

For example, the density of lines through A is greater than the density through B. Therefore, the electric field is larger at A than at B.A B

! E

! E (r) =

! F q"

=1q"

kQq"! r 2

ˆ r ⎛

⎝ ⎜

⎞

⎠ ⎟ = k

Qr2 ˆ r .

The field is radial and its magnitude depends only on ! r

(= r). For positive Q, the field lines are directed outward,

since ! F ( = q"

! E ) is directed radially outward and

! E ℓ

! F ,

when q" is positive. The number of lines leaving the

charge is proportional to Q, i.e., a charge of 2Q would have twice as many field lines.

Electric field lines for a single isolated positive charge:

++Q

q"

+

a

b

Fa > Fb

+

The electric field variation along the x-direction for a charge Q at the origin:

! E (x) = k

Qx2

ˆ i .

The direction of the electric field is determined by the direction of the force experienced by a positive test charge.

Ex

x

i +

+

+ve direction

x > 0 −ve direction

x < 0

+ q" q"

Q

EFM10VD1.MOV

Question 21.5: Why does the girl’s hair stand on end when she puts her hand on a Van der Graaf generator? NOTE: She is standing on an insulated table.

+

The girl’s body becomes charged ... so does her head and her hair. The strands of hair repel each other ... the most stable arrangement is

when her hair “stands on end”.

Because they are charged, the individual hairs follow closely the lines of the electric field produced by her head ... they act as “test charges”.

+ ++

+ ++ +

++++

++++

++++

++

+

+

Electric field due to two equal but opposite charges:

The electric field is a vector and so the electric field at any point due to a number of charges is the vector sum of the individual electric fields:

i.e., ! E (! r )= !

E ii∑ (! r ).Note in this example:

! E 1a =

! E 2a but

! E 1b >

! E 2b .

! E 1a

! E 1b

+Q −Q

! E 2b

! E b

! E a

! E 2a

a b

1 2

The number of lines leaving or entering a charge depends on the magnitude of the charge (we will prove this later). Since the magnitudes of the charges are the same, the same number of ‘lines’ leave the +ve charge (source) as enter the −ve charge (sink).

Electric field lines for two equal but opposite charges:

Fa > Fb > Fc

∴! E a >

! E b >

! E c

− Q + Q

+

+

+

a

b

c

The electric field variation along the x-direction for two charges ±Q, at x = ±a. The direction of the electric field is determined by the direction of the net force experienced by a positive test charge.

Ex

x i

+ −

+

−ve

x > a

+ve

a > x > 0

+ve

−a < x < 0

−ve

x < −a+

+

+

Question 21.6: Suppose we have two oppositely charged objects and we separate them farther. Has the energy of the system of charges changed or is it unchanged? Explain your answer.

+ −

The energy of the system of charges has increased. How do I

know that? Because I have to do positive work to separate the charges farther. Assume initially the charges are a distance a apart and I move them to a distance b apart. Then the work done is

W =

! F • d! x =

a

b∫ k

Q1Q2x2

a

b∫ dx

= kQ1Q2 −

1x

⎡ ⎣ ⎢

⎤ ⎦ ⎥ a

b= kQ1Q2

1a−

1b

⎛ ⎝ ⎜

⎞ ⎠ ⎟ > 0.

Where does the energy go? ... into the electric field!

After all an electric field can do work (by making charges move), so it must contain (potential) energy. In another chapter we will calculate the energy of an electric field.

Think ... what about gravitational and magnetic fields ... do they contain energy?

+ − Q1 Q2

a

b

Electric field lines for two positive charges with the same magnitude:

Since the magnitudes of the charges is the same, the same number of ‘lines’ leave each charge.

• What would be the force on a test charge at the point A, midway between the two charges ?

Q QA

! F 1

! F 2

! F

+ +

+

! E 2

! E 1

! E

+ + Q Q

A+

+

+

Question 21.7: Two point charges, q1 and q2 each with a

charge of +6.00nC, are placed on the y-axis at

y1 = +3.00cm and y2 = −3.00cm,

respectively.

(a) What are the magnitude and direction of the electric field on the x-axis at x = 4.00cm?

(b) What is the force exerted on a third charge

q3 = 2.00nC placed on the x-axis at x = 4.00cm?

(a) The resultant electric field: ! E R =

! E 1 +

! E 2.

Note that by symmetry: ! E 1,y +

! E 2,y = 0.

! E 1,x = k

6 ×10−9

(5×10−2)2 cosθ ˆ i = 1.73 ×104 ˆ i N/C

! E 2,x = k

6 ×10−9

(5×10−2)2 cosθ ˆ i = 1.73 ×104 ˆ i N/C

∴! E R = 3.46 ×104ˆ i N/C.

(b) Force on charge q3 = 2 nC:

∴! F = q3

! E R = 2 ×10−9 × 3.46 ×104 ˆ i

= 6.91×10−5ˆ i N

i

j 3 cm

3 cm 4 cm

! E 1

! E 2

5 cm

5 cm

θ

1

2

+

+

! E Rθ

+

The electric field along the x-direction variation for two charges +Q, at x = ±a. The direction of the electric field is determined by the direction of the net force experienced by a positive test charge.

Ex

x i

++

+ve x > a

−ve a > x > 0

+ve −a < x < 0

−ve x < −a

+

++

+

Question 21.8: (Review question.) Shown here are the electric field lines near two unlike charges of different magnitudes.

(a) What are the signs of the charge on Q1 and Q2?

(b) What is the ratio Q1

Q2?

(c) Some of the electric field lines from Q1 end on Q2.

Where do the electric field lines that do not end on Q2

go? Do they just stop?

Q1 Q2

(a) The electric field lines are outward from Q1 (so it’s

+ve) and inward towards Q2 (so it’s −ve).

(b) The number of electric field lines going into or out of a charge depends on the magnitude of the charge. The ratio of the number of lines leaving Q1 to the

number of lines going into Q2 is:

16

8 = Q1Q2

= 2, i.e., Q1 = 2Q2.

Q1 Q2

1 23456

78

9101112

13

14

1516

12

3 4

56

7

8

(c) No, they don’t stop! Electric field lines always begin on positive charge (the source) and end on negative charge (the sink). Therefore, the “extra” lines must end up on nearby negative charges, such as on you, or on the table, or on the walls!

In this example, a rubber rod ( −ve) has been rubbed by fur ( +ve). Note that many of the electric field lines end on the walls of the room!

Electric field for two oppositely charged plates (with total charge ±Q). Notice that the electric field lines are parallel and uniformly spaced except at the edges of the plates. So, the electric field between the plates is constant, which means that the force exerted on a charge is also constant between the plates.

We will derive the expression for the electric field between two charged plates in the next chapter.

++++++++

--------

! E

+Q −Q

+ ! F = q"

! E

Question 21.9: A charged particle leaves the origin with

a speed of 3.00 ×106 m/s at an angle of 35! above the x-

axis. A uniform electric field, " E = −E!ˆ j , exists

throughout the region. Find E! such that the particle will

cross the x-axis at x = 1.50cm if the particle is (a) an electron, and (b) is a proton.

(a) To deflect an electron downward, the field must be directed upward. We have ...

! v "x = 3 ×106 cos35" i = 2.46 ×106 ˆ i m/s.

! v "y = 3 ×106 sin 35" j = 1.72 ×106 j m/s.

The force on the electron is

! F = q

! E = −eEy j ,

i.e., it is directed only in the −y direction; there is no force (i.e., no acceleration) in the x direction, so v"x is

constant. (Does this remind you of something we’ve studied before?). Time connects the x and y motions. The time of flight from x = 0 to x = 1.50 cm is:

t =

Rangespeed

=1.5 ×10−2

v"x= 6.1×10−9s.

i j

x

y 35"

! v " ! E = Ey ˆ j

- 1.50 cm

i j

x

y 35!

" v ! " E = Ey ˆ j

- 1.50 cm

In the y-direction, we have:

ay =

Fm

= −eEym

and vy = v!y + ayt .

But, when the electron reaches the bottom plate, vy = −v!y,

∴ t = −

2v!yay

=2 v!y m

eEy= 6.1×10−9s.

i.e.,

Ey =

2 v!y m

et=

2 ×1.72 ×106 m/s × 9.1×10−31kg1.6 ×10−19C × 6.1×10−9s

= 3.21×103 N/C.

i j

x

y

(b) To deflect a proton downward, the field must be directed downward. We have again ...

! v "x = 3 ×106 cos35" i = 2.46 ×106 ˆ i m/s

! v "y = 3 ×106 sin 35" j = 1.72 ×106 j m/s

The force on the proton is

! F = q

! E = −eEy j ,

i.e., it is directed only in the −y direction; there is no force (i.e., no acceleration) in the x direction, so again

v"x is constant. Time connects the x and y motions. The

time of flight from x = 0 to x = 1.50 cm is:

t =

Rangespeed

=1.5 ×10−2

v"x= 6.1×10−9s,

which is the same as before.

35" ! v "

! E = Ey ˆ j

1.50 cm+

i j

x

y 35!

" v ! " E = Ey ˆ j

1.50 cm+

In the y-direction, we have:

ay =

Fm

= −eEym

and vy = v!y + ayt .

But, when the proton reaches the bottom plate, vy = −v!y,

∴ t = −

2v!yay

=2 v!y m

eEy= 6.1×10−9s.

i.e.,

Ey =

2 v!y m

et=

2 ×1.72 ×106 m/s ×1.67 ×10−27 kg1.6 ×10−19C × 6.1×10−9s

= 5.89 ×106 N/C.

Note that Ey ∝ m. Since mp > me, the electric field is

greater in the case of the proton by the ratio mp

me.

The electric dipole moment is defined as:

! p = q

! d

where ! d is the displacement from −q to +q. In a uniform

field, the forces ! F 1 and

! F 2 are equal but opposite and so,

they form a couple. The magnitude of the net torque on the dipole about the rotation axis is

τ = F1

ℓ2+ F2ℓ2= Fℓ = (qE)(d sin θ) = pEsinθ,

where θ is the angle between the dipole moment and the field direction. In this case the dipole will rotate clockwise.

NOTE: there is no net force acting on the dipole!

Electric dipoles ...

+

−

++q

! F 1 = q

! E

! F 2 = −q

! E −q

ℓ = dsin θ−

θ ! d

Potential energy of a dipole in a field ...

The work done by the field to rotate the dipole through dθ is:

dW = −τdθ = −pEsin θdθ. Therefore, the change in potential energy is:

dU = −dW = pEsin θdθ. So, the potential energy of the dipole is:

U = dU∫ = pE sin θdθ∫

= −pEcosθ = −! p •! E

• When θ = 0, U has its minimum value (stable equilibrium).

• When θ = 180", U has its maximum value

(unstable equilibrium).

The torque acts to align the dipole with ! p parallel to

! E .

0

180o 360o θ

U

θ

! E

! p

+

−

The water molecule is a POLAR MOLECULE with

! p = 6.17 ×10−30 C ⋅m.

That fact is the basis of the operation of a microwave oven.

The oscillating electric field within a microwave oven produces an oscillating torque on the water molecules that causes them to flip back-and-forth, producing heat. The heating effect is maximal at microwave frequencies (about 2.45 billion Hz).

+ −

− +

H H

O

+

−

− +

If a molecule is placed in an non-uniform electric field (say, due a single isolated positive charge), the molecule may become polarized, i.e., the positive and negative charge distort slightly forming a dipole. Since the negative charge is closer to the point charge than the positive charge ...

... there is a net attraction.

! F 1 >

! F 2

Why?

! F 1

! F 2+−

+

Question 21.10: What difference would it make if the central charge (shown in the previous slide) were negative instead of positive?

A: The molecule would be repelled from the central charge.B: The molecule would remain stationaryC: The molecule would be attracted to the central charge.

−

If the charge were negative, the electric field lines would be directed inward. The molecule would again be polarized but now, the positive end would be closer to the

central, negative charge. Again ! F 1 >

! F 2 so the molecule

would still be attracted to the charge. So, it makes no difference what the sign of the charge is, there is always attraction!

Therefore, the answer is C.

−

! F 2

! F 1

+−

The electric field of a positively charged plastic comb produces an attractive force on the molecules in a stream of water.

++ ++++++

H

HO + −

Question 21.11:

What would happen to a neutral (but polarizable) object if it were placed in a uniform field?

A: Nothing, it would remain stationary.B: It would move to the left.C: It would would to the right.D: It would rotate clockwise.E: It would rotate counter-clockwise.

The electric field exerts forces on the charges in the object so the molecules would become polarized, with the

−ve charges moving in the opposite direction to the field, i.e., to the left. The object becomes a dipole and would experience a torque that tends to turn it clockwise.

The answer is D ... it would rotate clockwise.

Note: because the sizes of the +ve and −ve charges are the same, there is no net force on the object, so there is no translational motion, only rotational motion!

+−