Phys 230 Winter 2010 - Chapter 21

Electric Charge and Electric FieldElectric charge, structure of matter, charge conservation and quantization

Conductors, insulators and induced charges

Coulomb‟s law and superposition principle

Electric fields and forces

Electric-field calculations

Point, dipole

line, ring

disk/planesphereElectric field lines

Electric dipoles

1 212 2

0

| |1

4

q qF

r

00

limq

FE

q

2

0

1ˆ

4

i

i i

qE

r r

The three big ideas in Chapters 21/22

Two basic problems in electrostatics:

Given a charge distribution

determine the electric field (and hence

force on a charge) at point P

OR

Given the electric field in a region

determine the charge distribution

1 2

122

0 12

1ˆ

4

q qF r

r

0 00

limq

FE

q

Coulomb‟s

Law

Electric

Fields

Gauss‟s

Law0

encqAdE

Electric Charge

Glass rods, plastic tubes, silk, and fur can

be used to demonstrate the existence of

“electrical forces” and two kinds of electrical

charge.

Today this is understood as due to the

movement of subatomic constituents called

electrons, whose presence or absence

lead to powerful forces of attraction and

repulsion.

Like charges repel and opposites attract.

Electrical charge is the quantum number that measures

the strength of the electrical force between objects.

Structure of Matter

How has rubbing the rod to „charge it‟ physically changed it?

Answer requires an understanding of atomic structure:

atoms consist of a dense core (nucleus) of protons and neutrons (whose dominant

interaction is via the strong nuclear force) occupying only one 10000th of the

atom’s volume, and relatively light discretized bundles of mass (electrons) that

statistically occupy the atom’s volume as a ‘probability cloud’.

In „bulk matter‟ the electrons are loosely bound to atoms/molecules/metal

lattices etc. (binding energy only a few eV often) via electrical forces, whereas

in stable matter the protons are very tightly bound in the nuclei (binding

energy on mega electron volts) via the strong force.

Protons and electrons carry one unit of an intrinsic property called electrical charge.

Structure of Matter

Thus it is the electrons (in solids1) that form a highly mobile carrier of electrical charge,

opposite of that of the proton. Thus:

A negatively charged object has a

surplus of electrons (relative to neutral)

A positively charged object has a

deficit of electrons (relative to neutral)

1In solutions, there are both positive and

negative charge carriers: in the form of ions.

Charging therefore refers to the transfer

of electrons from one object to another.

Notice implicitly, this does not involve a change in the total charge of the system.

Charge quantization

As we have seen, the matter is ultimately discrete: it is built up from fundamental, (essentially) indivisible

constituents. In particular therefore charge is not infinitely divisible, and there is a smallest (free) quantum of

charge experimentally – that carried by an electron (or proton):

e = 1.602×10-19 C

Every other charge is an integer multiple of this fundamental unit of charge. (Milikan oil drop expt.)

Charge quantization is a subtle theoretical issue (i.e. it‟s usually an input into the

theory, unless we are talking about certain Grand Unified Theories which predict

charge quantization).

Although charge is quantized, we will very often consider charge distributions which we approximate as

continuous – and hence can integrate over. We can do this because we are working in a classical approximation

where we have a very large number of fundamental charges and consider the motion/transfer of a relatively

small number of this total number..

The existence of quarks is a notable exception to this, but

quarks are confined into hadrons that have integer charges

Conservation of electrical charge:

The algebraic sum of all electrical charge in a closed system is a constant in time.

Electrical charge can neither be created nor destroyed.

OR

Charge Conservation

In any charging or discharging process charge is merely transferred from one place to another.

The conservation law is a universal one – no exceptions have ever (or likely will ever) be observed. Why?

Like energy, momentum and angular momentum conservation, charge conservation is tied to a symmetry of

nature. Whereas the former three are intimately tied to spacetime symmetries, charge conservation is tied

to an abstract, “internal” phase rotation symmetry symmetry:ie (Noether‟s theorem)

Conductors versus Insulators

Since matter is electrically neutral in bulk, we will often think of charges moving within matter.

Three scenarios we will consider:

1) Stationary charges: v = 0 (electrostatics)

2) Charges moving uniformly: v = const (magnetostatics)

3) Charges accelerating: v changes (electrodynamics)

Conductors: materials in which (valence) electrons are relatively free to move, in a sea of loosely bound electrons.

Think metals.

Insulators: materials which have few available “conduction band” electrons. Think non-metals.

Charging by Conduction

The process of giving one object a net charge by placing it in contact with another object.

Charging by Induction

The process of charging a second object by placing a charged object near it. In a conductor the free electrons are

attracted or repelled by the presence of the nearby charge.

The ground effectively acts as an infinite reservoir to absorb extra charge.

In an insulator, neutral molecules can realign themselves so that the opposite charges are closer to the

nearby charge (polarization), leading to an attractive force between the charged object and the neutral one.

Charging by Induction – Polarization/Electrical forces on uncharged objects

Coulomb`s Law

The basic question in electrostatics: given a configuration of stationary charges, what force acts on a test charge?

We are now interested in a quantitative description of these electrical forces. In fact...

The basic tool is Coulomb`s law:

The magnitude of the electrostatic force between two point charges q1 and q2 is proportional to

the size of the charges and inversely proportional to the square of the distance between them:

1 1 22

2

1

1

0 2

2 2

12

1

4

q qF k

r

q q

r 9 2 2

12 2 2

0

8.988 10 N m / C

8.854 10 C / N m

k

++–

The force is attractive if sgn(q1)= – sgn(q2 ), and repulsive if sgn(q1)=sgn(q2) ,

and lies along the line that connects the two points.

YF21.82 (Electric Forces inside the nucleus) Typical dimensions of atomic nuclei are of the order 10-15 m (1 fm).

(a) If two protons in a nucleus are 2.0 fm apart, find the magnitude of the electric force each one exerts on the

other. Could you feel this force?

(b) Since the protons repel each other so strongly, why don‟t they shoot out of the nucleus?

Moral: If you see a Coulomb of charge in a dark alley... RUN!

Coulomb`s Law – Direction and Superposition principle +–

1

212r̂1 212 122

0 12

1ˆ

4

q qF r

r

12r̂ = unit vector pointing from the pos. of 1 to that of 2

In vector form:

Note Newton‟s third law implies: 21 12F F

How do we compute the force on a given point charge Q due to the presence of multiple point charges?

r2

r1

r3

q1

q2q3

“1 on 2”

Linear superposition principle: the net force on a point charge

Q from a set of point charges q1, q2, q3, ... is given by the

pairwise VECTOR sum of the forces due to each charge

Q

2

0

ˆ4

iQ i

i i

qQF r

r VECTOR sum

2

0

ˆ4

Q

Q dqF r

r

For a static continuous charge distribution, the force on a point charge Q

is given by the integral version of this:

but it is more useful to define the electric field for continuous

distributions (next section)

Intuitive meaning: Interaction between two charges

completely unaffected by presence of other charges.

1D Example (YF21.20)

Two point charges are placed on the x-axis as follows: charge q1 = +4.00 nC is located

at x = 0.200 m, and charge q2= + 5.00 nC is at x = –0.300 m. What are the magnitude

and direction of the total force exerted by these two charges on a negative point

charge q3 = – 0.600 nC that is placed at the origin?

2D Example (YF21.78)

Two point charges q1 and q2 are held in place 4.50 cm apart. Another point

charge Q = – 1.75 μC of mass 5.00 g is initially located 3.00 cm from each of

these charges and released from rest. You observe that the initial acceleration

of Q is 324 m/s2 upward, parallel to the line connecting the two point charges.

Find q1 and q2.

YF21.74 Two identical spheres with mass m are hung from silk threads of length L as

shown. Each sphere has the same charge, so q1=q2=q. The radius of each sphere is

very small compared to the distance between the spheres, so they may be treated as

point charges. Show that if the angle θ is small the equilibrium separation

between the spheres is 1/32

02

q Ld

mg

YF21.73 Two positive point charges Q are held fixed on the x-axis at x=a and x = –a

A third positive point charge q with mass m is placed on the x-axis away from the origin

at a coordinate x such that . The charge q, which is free to move along the

x-axis is then released. (a) Find the frequency of oscillation of the charge q. (b)

What happens if the charge is placed on the y-axis instead?

multiple concept questions

with math approximations

2(1 ) 1 ( 1) / 2 ...nz nz n n z

ax

ay

Taylor series/Binomial Expanision:

Electric Field

Since the electrostatic force between charges exists even over empty space,

how does one charge ‘know’ about the other charge?

There is something extremely disturbing the interaction picture we‟ve painted thus far:

The same issue arises

in Newtonian gravity!

“Action at a distance”

OLD PERSPECTIVE

charge charge

Forbidden by relativity

NEW(er)

Field as intermediary charge chargefield

Each point charge, by virtue of its existence, creates a „field‟ around it in all directions which „instructs‟ other

charges how to behave when immersed in the field. In other words, charges modify the space around them.

Contemporary

chargesea of virtual

quanta (photons)charge

“quantum field”

generates

interact directly

with other

(charges directly interact

over empty space)

Electric Field To elaborate...

The choice of P was arbitrary, and so the electric field (a „vector field‟) exists

everywhere but at A itself where it is not defined.

Changes in the electric field (say due to a moving source distribution) are propagated away from the source at

the speed of light. For now we are interested in static charge distributions for which the electric field is fixed.

The electric field at P yields the force per unit charge on a small test charge placed at P. It depends on the size

and locations of the source charge(s) that set up the field, but NOT the test charge. Furthermore, the field does not

exert a force on the source charges – there are no self-interactions! (This is essentially part of Newton‟s 2nd law.)

0 00

limq

FE

qTechnically the E field at P is the formal limit ,

since we don‟t want to consider the effect of the

test charge itself on the source charges and hence the

electric field.

Electric Field of a Point ChargeThe simplest charge configuration is an isolated point charge Q. Its electric field is given

by Coulomb‟s law:

020

0

ˆ( ) limq

rr

F kQE r

q

where is the radial unit vector

pointing away from the charge Q. r̂

The E-field viewed as a

vector field (i.e. a vector assigned

to each point in space) about the

point charge therefore looks like...

Notice the spherical symmetry: E only depends

on r, the radial distance from the point charge.

An Analogy with Gravity

Actually you‟ve been working with a vector field as long as you‟ve been studying physics...

Consider Newton‟s gravitational law:2

ˆeGM mF r mg

r

2( ) eGM

g rr

What you‟ve been calling the „gravitational acceleration‟ of the Earth is really the „gravitational field‟ of Earth:

the force per unit mass experienced by a test mass placed near the Earth.

Here „mass‟ plays the role of „gravitational charge‟ (it‟s always attractive).

The gravitational field near the earth’s surface is constant in magnitude (since RE is so large), but always

points towards the center of the Earth: an Australian and a Canadian disagree on the direction of the

gravitational field even though they both agree all objects accelerate at 9.8 m/s2.

vs. 2( )

kQE r

r

We‟re being sloppy since the Earth

is not a point mass, but the shell

theorem will reconcile this.

Electric field of an Electric Dipole (Example 2) – Superposition/Vector addition

An electric dipole consists of two charges of magnitude Q, but opposite in sign held a small distance d apart. A

„long way‟ from the dipole the electric field due to the opposing charges tends to cancel and we expect a

E-field fall-off faster than 1/r2 (after all, if the charges precisely coincided, the net charge would be zero, and there

would be no electric field anywhere).

+

–

(See also example 21.9)

Find the E-field along the line perpendicular to the line joining the charges through the midpoint.

d

Px

3 2 2 3/2

ˆ 1

(1 / 4 )P

kQd jE

x d x

Result:

Large x limit:

3

ˆP

kQd jE

x

cubic falloff

characteristic

of dipoles

Electric field of an Electric Dipole cont. (See also example 21.15)

+ –d P

xRepeat the calculation for a point far along the axis of the dipole...

2 2(1,0) ( 1,0)

( / 2) ( / 2)P

kQ kQE E E

x d x d

x=0

Taylor/Binomial expand:2(1 ) 1 ( 1) / 2 ...nz nz n n z

Result:

In both cases we‟ve considered, the combination Qd arises (in addition to the cubic fall-off). This is called the

(magnitude of the) dipole moment, and is denoted by p. Its direction is defined to point from the negative

charge to the positive charge:

for

p QdWe will return to dipoles at the end of this chapter.

1 zdx

ix

kQdEP

ˆ2~

3

dx

Electric Field Lines

Before we compute the electric field due to more complicated charge configurations, let us consider the

visualization of (otherwise abstract) electric fields a little further.

An electric field line is a curve whose tangent at any point is in

the direction of the electric-field vector there.

Faraday‟s Insight: “Connect the dots”, or in this case connect the arrows

By connecting the arrows, we don‟t lose information about the magnitude of the E –field: instead it is encoded by

the „density of the field lines‟: denser regions correspond to stronger fields.

Warning: field lines are lines of force, NOT trajectories. E-fields give us forces and hence accelerations,

NOT velocities. (Recall EnPh 131: the velocity vector is tangent to trajectory, not the acceleration.)

Electric Field Lines – Other notes

Field lines, like particle streamlines in a

fluid never cross: the E-field is unique at

each point.

By defn, field lines start on positive charges

and end at negative charges or at infinity.

The field lines themselves are unphysical,

though we can get polar substances to

„line up‟ along them.

Electric Field Calcuations: Continuous Charge Distributions

The E-field for a discrete set of point charges is computed from the superposition principle: 2

0

1ˆ

4

iP i

i i

qE r

r

Often we are concerned with computing the E-field due to a continuous charge distribution described in

1D by a linear charge density: / dxdq charge per unit length (as a function of x, y and z)

2D by a surface charge density:

The summation in the discrete case becomes an integral in the continuous case:

/ dAdq charge per unit area (as a function of x, y and z)

The hard part is to express the differential element of charge dq in terms of the geometry.

3D by a volume charge density: / dVdq charge per unit volume (as a function of x, y and z)

2

0

ˆ1

4P

Cr

dxE

r

e.g.

It is also crucial to note that this is a vector integration. We‟ll now look at several examples.

You need to learn these techniques well: the same methods will be used again in the calculation of magnetic fields.

E-field: Uniform line charge along perpendicular bisector (Example 1)L/2

– L/2

dqConsider a charge Q uniformly distributed along a rod of length L lying on the

y-axis from y = – L/2 to y = L/2. Compute the E-field at a point along the x-axis.

Uniformdq Q

constdy L

Differential element: /dq dy Qdy L

Differential element contribution to E-field at P: 2

0

1(cos , sin )

4

dydE

r

E-field at P:/2

2 2 3/2 2 2/20 0

ˆ ˆ ˆ 1

4 ( ) 2 1 4 /

L

L

xi yj iE dE dy

x y x x L

Long line/short distance limit:( / 0)x L 0

ˆ

2

iE

x

1/r falloffLong distance limit:

( / )x L 0

2

ˆ

4

QiE

x

1D

(y-comp. has

odd integrand)

SEVERE TIRE DAMAGE WARNING:

If the charge was NOT UNIFORMLY distributed along the line, so that then it is false and

in fact meaningless to assert since λ is explicitly a function of y while the rhs is a constant.

In this (commonly occurring in this course) case, you must leave and integrate using

the given functional form of λ.

)( y

LQ /

dyydq )(

E-field: Uniform ring of charge, along its symmetry axis (Example 2)

Here dE has x,y and z components depending on the line element.

But by symmetry1 the y and z components of opposite pairs of elements

on the ring cancel. Only the x-component is additive.

Differential analysis:2

dq Qconst

ds a

ds a d

dq

angular variable

around ring

x-comp. of differential element contribution to E-field at P: 2 2 2 3/2

0 0

1 1cos

4 4 ( )x

dq x a ddE

r x a

Total E-field: 2 2 3/2 2 2 3/2

0 0

ˆ ˆ

2 ( ) 4 ( )

a xi Q xiE

x a x a

(Note: x is constant in the integration!)

Small x/a limit:

Large x/a limit:

3

0

ˆ

4

QiE x

a

2

0

ˆ 1

4

QiE

x

(so –q, small x exhibits SHM)

(ring appears as point

charge from far away)

1D

1You can certainly set up the y and z components of the E-field (as we will do in

the equivalent magnetic field question), and then show explicitly that when you

integrate over the whole ring, the y and z components integrate to zero.

E-field: Uniform disk, built from infinitesimal rings, along symmetry axis (Example 3) 2D

Differential element: 2

dq Qconst

dA R

2dA daa

circumference thickness P

2

2Qdq a da

R,

E-field at P due to ring of radius a:(from previous slide!)

2 2 3/2

0

ˆ

4 ( )

dq xidE

x a

Now integrate over all rings: 2 2 2 3/2 2 200 0

ˆ ˆ1

2 ( ) 2

RQxi ada i xE

R x a x R

Close to disk

or large disk:( / 0)x R 0

ˆ

2

iE const

Long distance limit:

( / )x R(i.e. point charge)

Use previous result: dq is now the infinitesimal ring of charge, and we „build up‟ the disk by

summing (i.e. integrating) the contributions from the rings, by allowing the radius „a‟ of the

ring to vary:

x

2

0

2

2

0 4...

211

2

ˆ

x

Q

x

RiE

Again, the claim that σ = Q/πR2, holds and is meaningful

ONLY for a uniform charge distribution. If σ is NOT constant,

then you MUST work with dq = σ dA and the given functional

form of σ. Stay tuned for a nonuniform example!

E-field: Uniform disk cont. (parallel plate capacitor: example 21.13)

Since we have a whole chapter devoted to capacitance, it is worth investigating this example further. We have

found a configuration that yields a uniform E-field in the large disk/short-distance to the disk limit R→∞ or x→0).

In this limit, it does NOT matter that the disk is circular or that we remain in the center of it.

Now consider two oppositely charged closely spaced „infinitely large‟ (i.e. d/R→0) „plates‟:

1 2

0

0

ˆ

0

E E E j

above upper sheet

between the sheets

below lower sheet

superposition

A Nonuniform Example – Phys 230 Fall 2009 Midterm (Long Answer 1, parts b, c, and e partial)

A non-uniformly charged insulated disk of radius R lying in the yz plane with its center at the origin at x=0, has a

surface charge density given by:( )

Cr

r

b) If the total charge on the disk is Q, determine the constant C.

c) Use the E-field due to a ring and b) to determine the E-field due to this charged disk at any point on the

positive x-axis. Use the substitution r = x tan θ. (Symmetry dictates the E-field will still only have an x-component

on the x-axis.)

e) Verify the result in c) reduces to that of a point charge for x >> R.

where r is the distance from the center of the disk to a point in the yz plane.

Again, the idea is to build up the disk from infinitesimal rings over which sigma is approximately constant!

(„r‟ is what we were calling „a‟ in the previous

two examples. It‟s a more natural choice here,

since we want to emphasize that it is a variable.)

E-field: Uniform Spherical Shell “The Shell Theorem” (Example 4) 2D

2 2.

GMm kQqvs

r r

Newton invented calculus to solve this problem: what is the gravitational

force exerted on a point-like body due to a uniform spherical shell?

We‟re interested in the same problem, applied to electrostatics instead

of gravity: both are inverse-square law forces.

Differential element: (spherical ring) 2 sin RRdA d

circumference thickness

This is a another crucial application of our ring result.

E-field contribution from ring:2 3/2

0

24 ( )x

x

x a

dqdE

24

Qdq dA dA

R

r

Ra=R sin θ

θ

s

3

0

cos sin

8

sQ

s

d

cosx s

2 2 2 2 cosR s r rs 2 2 2 2 coss r R rR

Cosine law:

Spherical shell built

from thin rings

x = distance to ring center

r = distance to center of sphere

s = distance to ring surface

a &R = radius of ring & sphere

E-field: “The Shell Theorem” cont.

R and r are constant in the integration, but s, θ and are variable.

But they are not independent of each other as the animation shows...

2 2 2 2 cosR s r rs

2 2 2 2 coss r R rR

Use the cosine law twice to eliminate the angles in favour of s:2 2 2

cos2

s r R

rs

sins ds rR d differentiateThus:

2 2 2

2 2

016x

Q s r RdE ds

r R s

This easy integral evaluates to:

2 22

02 2| |0

4116

0

r R

xr R

Qr RQ r R

rE dsr R s

r R

(Shell seen as a point

charge from outside!)

(No E-field inside shell!)

Shell theorem geometry

(labelled for gravity q→m)

E-field: “The Shell Theorem” Discussion

Given the spherical symmetry, the choice of the

“x-axis” is irrelevant, and the result is general:

2

0

1ˆ

4

QE r

r anywhere outside the shell

0E anywhere inside the shell

Furthermore, with no further work, we can conclude for a uniformly charged solid sphere/ball, the result is the

same. Why? Because we can build the uniform ball up from a series of concentric shells... the shell result tells

us the E-field (outside) depends only on the distance to the center. But each concentric shell has the same center!

This is a very important and simplifying result (in both electrostatics and Newtonian gravity) that we‟ve implicitly

been using all along: it allows us to treat certain objects as points.

As we have seen, symmetry principles play an important role in simplifying many calculations. In the next

chapter we will introduce Gauss’s Law which will allow us to obtain this result in a much simpler manner.

YF21.99 (Example 5: Composite configurations)

Two 1.20 m non-conducting wires meet at a right angle. One segment

carries +2.50 μC of charge distributed uniformly along its length, and the

other carries – 2.50 μC of charge distributed uniformly along it, as shown.

(a) Find the magnitude and the direction of the electric field these wires

produce at point P, which is 60.0 cm from each wire.

(b) If an electron is released at P, what are the magnitude and direction

of the net force that these wires exert on it?

(Use earlier results! Note the rods are not infinite.)

Important special case: Uniform E-fields and Conductors

We have seen that close to a large uniformly charged plate, or between two large uniformly charged plates, the

E-field is constant in both magnitude and direction.

Another case of importance in which a uniform E-field arises is in the INTERIOR of a CONDUCTOR in an

ELECTROSTATIC situation: namely the E-field is identically ZERO.

Why?

Memorize this fact.

Consider the contrapositive: Suppose the E-field in the interior of a conductor in electrostatic equilibrium was

non-zero. Since it is a conductor, charges on the interior would be free to move under the influence of the non-zero

field. But if they were free to move (and in particular accelerate), then this would not be an electrostatic situation.

Thus the E-field in the interior must be zero, and any excess charge on the conductor must be at its surface(s).

(Note this argument says nothing about the E-field in a hole inside a conductor.)

Behaviour of charges in a uniform field

Our discussion so far has been centered around computing the E-field due to a given static charge distribution.

The „other side‟ is to use the E-field to compute motion of a small1 charge placed in a given E-field. The problem

is that most E-field configurations yield mathematically difficult/impossible to compute trajectories.

1The restriction to small charges is so that we may neglect the

„back-reaction‟ of the charge on the E-field itself!

However one easy case to deal with is the behaviour of charges immersed in uniform E-fields.

uniform 1D

acceleration

essentially 2D

projectile motion

YF 21.87 (Motion through a Uniform E-field example) - Partial

A proton is projected into a uniform electric field that points vertically

upward and has magnitude E. The initial velocity of the proton has a

magnitude v0 and is directed at an angle α below the horizontal.

(a) Find the maximum distance hmax that the proton

descends vertically below its initial elevation. (Ignore gravity.)

(b) After what horizontal distance d does the proton return to its

original elevation?

(c) Sketch the trajectory of the proton.

E0v

Dipoles revisited: dipoles immersed in uniform fields

We`ve discussed the E-fields created by an electric dipole, and we conclude

the chapter with a discussion of the behaviour of dipoles in an external field.

You owe your existence to

the polar nature of water:

without it biochemistry is

impossible.

Quantum mechanics

explains the polar

nature of water.

Since the total charge on a dipole is zero, the net

force acting on the rigid dipole is zero. However

since the forces that act individually on the charges

act along different lines, the net torque acting on a

dipole is not zero. With respect to the center of the

dipole:sin sinqEd pE

is the angle between p and E

Recall: p is the dipole moment. It

points from the –ve to the +ve charge.

p E

The minus sign is to indicate explicitly the

torque is clockwise and acts to decrease phi.

Dipoles revisited: dipoles immersed in uniform fields

Thus the torque is maximal when the dipole is perp. to the E-field, and zero when they are parallel or antiparallel.

The external E-field tends to align the dipole so that the dipole moment p is parallel to the E-field (stable

equilibrium, why?,YF21.69). (The anti-parallel configuration is therefore unstable.)

Work done on a dipole by a uniform field

Recall from EnPh131 the work done by a force is given by: W F ds

The electrostatic force does work in rotating an out-of-equilibrium dipole to its equilibrium position. Furthermore,

like gravity, electrostatic forces are conservative, so it makes sense to speak of a potential energy of a dipole...

The rotational analogue of this is given by applying our dictionary: W d (or parametrize the path

element by: and

note that .)

ds Rd

tanF R

Dipoles revisited: Potential energy in uniform fields

Recall, by definition:0 0

0 0sin cos cosf f

f C fU U U W d pE d pE pE

Thus adopting the universal convention that

( ) cosU pE p E

For small angular displacements

about stable equilibrium: 2 / 2

pE

U pE

(0)U pE

versus a spring:2 / 2

F kx

U kx

or a pendulum:

2 / 2

mgl

U mgl

YF21.70 A dipole consisting of charges ±e, 220 nm apart is placed between two very large sheets carrying equal

but opposite charge densities of 125 μC/m2. (a) What is the max potential energy this dipole can have due to the

sheets? (b) What is the maximum torque the sheets can exert on the dipole? Orientations for a) and b)?

Remember only changes in potential

energy have meaning. The „zero‟ location

is arbitrary.

(Supplementary)

Chapter Review Questions – Phys 230 Fall 2009 Midterm (Questions 1 and 2)

Multiple Choice/Short Answer: Where indicated, you must show the summary of your work (key concepts or eqns)

tailored specifically to the question in order to obtain ANY marks (i.e. no random guessing). Keep it brief (your details can

be done on scrap); I don’t want to mark these as long answer questions!

1. (4 marks) A spherical conductor of radius r = 50.0 cm and carrying a charge of +4.00 μC is centered at the origin. An

infinite line charge of uniform linear charge density λ = – 2.00 μC/m runs parallel to the x-axis at y = 3.00 m. What is the

magnitude of the E-field at the point P on the x-axis 2.00 m to the right of the origin?

P

4 μC

–2 μC/m

3.0 m

2.0 m

x

y

a) 3.00×103 N/C

b) 1.08×104 N/C

c) 1.50×104 N/C

d) 2.00×104 N/C

e) 2.10×104 N/C

Formula(e) Used: _______________________________________________________________

2. (4 marks) Two vertically-oriented oppositely-charged large metal plates are placed a distance 0.25 m apart, and the

potential difference between them is 70 V. A very small metal ball with a mass 8.0×10-4 kg, and a charge of +5.0 μC is

attached to an insulating thread, and lowered into the space between the plates. What angle will the thread make with the

vertical when the metal ball is equilibrium? (Use g = 9.81 m/s2.)

θ

+ –

a) 10°

b) 15°

c) 20°

d) 25°

e) 30°

Formula(e) Used: _______________________________________________________________