• Electrostatic Phenomena• Coulomb’s Law

12212

2112 ˆ

4

1r

r

qqF

ope

=r

q1• Superposition

qFtotal = F1 + F2 + ...q2

F1

F2

F

Two balls of equal mass are suspended from the ceiling withnonconducting wire. One ball is given a charge +3q and the other is given a charge +q.

g+q+3q

Which of the following best represents the equilibrium positions?

(b)

+3q+q

(a)

+3q+q

(c)

+3q +q

+3q

(b)

+3q+q

(a)

+3q+q

(c)

+q

Which best represents the equilibrium position?

• Remember Newton’s Third Law!• The force on the +3q charge due to the +q charge must be equal

and opposite to the force of the +3q charge on the +q charge• Amount of charge on each ball determines the magnitude of the

force, but each ball experiences the same magnitude of force• Symmetry, therefore, demands (c)

P.S. Knowing the form of Coulomb’s law you can write two equations with two unknowns (T and q )

Two charges q = + 1 μC and Q = +10 μC are placed near each other as shown in the figure.

6) Which of the following diagrams best describes the forces acting on the charges:

Preflight 2:

+10 μC+1 μC

a)

b)

c)

The Electric Field

•A simple, yet profound observation- The net Coulomb force on a given charge is alwaysproportional to the strength of that charge.

q

q1

q2

F1

F

F2

F2F1F = +

÷÷ø

öççè

æ+=

22

222

1

11

0

ˆˆ

4 r

rq

r

rqqF

per

test charge

- We can now define a quantity, the electric field, whichis independent of the test charge, q, and depends only onposition in space:

q

FE

rrº The qi are the sources

of the electric field

The Electric Field

q

FE

rrº

With this concept, we can “map” the electric fieldanywhere in space produced by any arbitrary:

“Net” E at origin

+

F

These charges or this charge distribution“source” the electric field throughout space

ii

i rr

qE ˆ

4

12

0

++

+ +

+

+-

-

--

-

Bunch of Charges

rr

dqE ˆ

4

12

0

+ ++ ++ +++++ +

Charge Distribution

Example: Electric Field

The x and y components of the field at (0,0) are:

1) Notice that the fields from the top-rightand bottom left cancel at the origin?

2) The electric field, then, is just the fieldfrom the top -left charge. It points awayfrom the top-left charge as shown.3) Magnitude of E-field at the origin is:

E2a2

kq=

What is the electric field at the origin due to this set of charges?

x

y

+q+q

+q

a

a

a

a a

2a

Q

Ex 2a2

kq= cosq

2

12a2

kq=

Ey 2a2

kq= sinq-

2

12a2

kq= -

Example: Electric Field

222 a

qQkQEF xx

222 a

qQkQEF yy

Now, a charge, Q, is placed at the origin. What is the net force on that charge?

+q

x

y

+q

+q

a

a

a

a a

2a

Q

Note: If the charge Q is positive, the force will be in the direction of the electric fieldIf the charge Q is negative, the force will be against the direction of the electric field

F is

F is

2

1

2 2a

qkEy

2

1

2 2a

qEx =k

Let’s Try Some Numbers...

2

If q = 5mC, a = 5cm, and Q = 15mC.

Then Ex = 6.364 106 N/C

and Ey = -6.364 106 N/C

Fx= 95.5 N Fy= -95.5 N

Fx = QEx and Fy = QEy

So... and

We also know that the magnitude ofE = 9.00 106 N/C

We can, therefore, calculate the magnitude of F

F = |Q| E = 135 N

x

y

+q+q

+q

a

a

a

a a

2a

Two charges, Q1 and Q2, fixed along the x-axis asshown produce an electric field, E, at a point(x,y)=(0,d) which is directed along the negativey-axis.

Q2Q1 x

y

Ed- Which of the following is true?

(a) Both charges Q1 and Q2 are positive

(b) Both charges Q1 and Q2 are negative

(c) The charges Q1 and Q2 have opposite signs

Two charges, Q1 and Q2, fixed along the x-axis asshown produce an electric field, E, at a point(x,y)=(0,d) which is directed along the negativey-axis.

- Which of the following is true?

E

Q2Q1

(a)

Q2Q1

(c)

E

Q2Q1

(b)

E

Q2Q1 x

y

Ed

(a) Both charges Q1 and Q2 are positive

(b) Both charges Q1 and Q2 are negative

(c) The charges Q1 and Q2 have opposite signs

Ways to Visualize the E FieldConsider the E-field of a positive point charge at the origin

+

+ chg

field lines

+ chg

vector map

+

Rules for Vector Maps

• Direction of arrow indicates direction of field• Length of arrows µ local magnitude of E

+ chg

+

• Lines leave (+) charges and return to (-) charges• Number of lines leaving/entering charge µ amount

of charge• Tangent of line = direction of E• Local density of field lines µ local magnitude of E• Field at two white dots differs by a factor of 4 since r

differs by a factor of 2• Local density of field lines also differs by a factor of 4

(in 3D)

Rules for Field Lines

+ -

3

6) A negative charge is placed in a region of electric fieldas shown in the picture. Which way does it move ?

a) up c) left e) it doesn't move b) down d) right

Preflight 2:

7) Compare the field strengths at points A and B.

a) EA > EB

b) EA = EB

c) EA < EB

• Consider a dipole (2 separated equal and opposite charges) with the y-axis as shown.

+Q

x

a

a-Q

a

2a

y

- Which of the following statements about Ex(2a,a) is true?

(a) Ex(2a,a) < 0 (b) Ex(2a,a) = 0 (c) Ex(2a,a) > 0

Ex Solution: Draw some field lines according to our rules.

• Consider a dipole (2 separated equal and opposite charges) with the y-axis as shown.

+Q

x

a

a-Q

a

2a

y

- Which of the following statements about Ex(2a,a) is true?

(a) Ex(2a,a) < 0 (b) Ex(2a,a) = 0 (c) Ex(2a,a) > 0

Two equal, but opposite charges are placed on the x axis. The positive charge is placed at x = -5 m and the negative charge is placed at x = +5m as shown in the figure above.

3) What is the direction of the electric field at point A?

a) up b) down c) left d) right e) zero

4) What is the direction of the electric field at point B?

a) up b) down c) left d) right e) zero

Preflight 2:

Field Lines From Two Opposite ChargesDipole

Dipoles are central to our existence!

Molecular Force Model Basis of Attraction

Ion-dipole Ion and polar molecule

Dipole-dipole Partial charges of polar molecules

London dispersion

Induced dipoles of polarizable molecules

The Electric Dipolesee the appendix for further information

q x

y

a

a

+Q

rE E

-Q

What is the E-field generated bythis arrangement of charges?

Calculate for a point along x-axis: (x, 0)

Ex = ??

Symmetry

Ex(x,0) = 0

Ey = ??

sin4

120,

20 r

QxEy

222 axr r

asin

2/32204

120,

ax

QaxEy

Electric Dipole Field Lines

• Lines leave positive chargeand return to negative charge

What can we observe about E?

• Ex(x,0) = 0

• Ex(0,y) = 0

• Field largest in space between two charges

• We derived:

... for r >> a,

3

1)0,(

xxEy

2/32204

120,

ax

QaxEy

Field Lines From Two Like Charges

• There is a zero halfwaybetween the two charges

• r >> a: looks like the fieldof point charge (+2q) at origin

4

• Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length.

• The electric field at the origin is

R

x

y++++

+ +

+++

++

+ + + + ++++++

+

(a) zeroR

2

4

1

0

(b) 204

1

R

R

(c)

• Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length.

• The electric field at the origin is

R

x

y++++

+ +

+++

++

+ + + + ++++++

+

(a) zeroR

2

4

1

0

(b) 204

1

R

R

(c)

• The key thing to remember here is that the total field at the origin is given by the VECTOR SUM of the contributions from all bits of charge.

• If the total field were given by the ALGEBRAIC SUM, then (b) would be correct. (exercise for the student).

• Note that the electric field at the origin produced by one bit of charge is exactly cancelled by that produced by the bit of charge diametrically opposite!!

• Therefore, the VECTOR SUM of all these contributions is ZERO!!

Electric Field inside a Conductor

• A two electron atom, e.g., Ca– heavy ion core– two valence electrons

• An array of these atoms– microscopically crystalline– ions are immobile– electrons can move easily

• Viewed macroscopically:– neutral

There is never a net electric field inside a conductor – the free charges always

move to exactly cancel it out.

2+

2+ 2+ 2+ 2+

2+ 2+ 2+ 2+

2+ 2+ 2+ 2+

2+ 2+ 2+ 2+

Summary

• Define E-Field in terms of forceon “test charge”

• How to think about fields

• Electric Field Lines

• Example Calculation: Electric Dipole

Appendix A:Other ways to Visualize the E Field

Consider a point charge at the origin

+

+ chg

Field LinesEx, Ey, Ez as a function of (x, y, z)Er, Eq, Ef as a function of (r, q, f)

Graphs

x

Ex(x,0,0)

Appendix A- “ACT”

(a) (c)(b)

f0 2p

Er3A

f0 2p

Er

f0 2p

Er

Fixedr > 0

3B

Fixedr > 0

f0 2p

Ex

f0 2p

Ex

f0 2p

Ex

Consider a point charge fixed at the origin ofa coordinate system as shown.

–Which of the following graphs bestrepresent the functional dependence ofthe Electric Field for fixed radius r?

xQ

y

r

f

Appendix A “ACT”Consider a point charge fixed at the origin ofa coordinate system as shown.

– Which of the following graphs bestrepresent the functional dependence ofthe Electric Field for fixed radius r?

xQ

y

r

f

(a) (c)(b)

f0 2p

Er3A

f0 2p

Er

f0 2p

Er

Fixedr > 0

• At fixed r, the radial component of the field is a constant, independent of f!!

• For r>0, this constant is > 0. (note: the azimuthal component Ef is, however, zero)

Appendix A “ACT” Consider a point charge fixed at the origin ofa coordinate system as shown.

–Which of the following graphs bestrepresent the functional dependence ofthe Electric Field for fixed radius r?

xQ

y

r

f

(a) (c)(b)

3B

Fixedr > 0

f0 2p

Ex

f0 2p

Ex

f0 2p

Ex

cosrx EE • At fixed r, the horizontal component of the field Ex is given by:

Appendix B: Electric Dipole

What is the E-field generated bythis arrangement of charges?

Calculate for a point along x-axis: (x, 0)

Ex = ??

Symmetry

Ex(x,0) = 0

Ey = ??

sin4

120,

20 r

QxEy

222 axr r

asin

2/32204

120,

ax

QaxEy

q x

y

a

a

+Q

rE E

-Q

Electric Dipole

x

y

a

a

+Q

-Q

0,0 yEx

Coulomb ForceRadial

22

0

11

4,0

ayay

QyEy

2

2

240

1

4

4,0

ya

y

ayQyEy

E

Now calculate for a pt along y-axis: (0,y)

Ex = ?? Ey = ??

What is the Electric Field generated by this charge arrangement?

Electric Dipole

x

y

a

+Q

-Q

Case of special interest:(antennas, molecules)

r > > a

r

For pts along x-axis: For pts along y-axis:

2/32204

120,

ar

QarEy

00, rEx

3

04

120,

r

QarEy

For r >>a,

E rx 0 0,

2

2

240

1

4

4,0

ra

r

arQrEy

For r >>a

3

04

14,0

r

QarEy

a