Example Problem

A particle with charge +2 nC (1 nanoCoulomb=10-9 C) is located at the origin. What is the electric field due to this particle at a location <-0.2,-0.2,-0.2> m?

rSolution:1. Distance and direction:

Example Problem

𝑟=⟨𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 ⟩− ⟨𝑠𝑜𝑢𝑟𝑐𝑒𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 ⟩𝑟=⟨ −0 . 2 , −0 . 2 , −0 . 2 ⟩ − ⟨ 0 , 0 , 0 ⟩

m

�̂�= 𝑟|𝑟|

=⟨− 0 .2 , − 0 .2 , − 0 .2 ⟩

0 . 35= ⟨− 0 .57 ,− 0 . 57 , −0 . 57 ⟩

𝐸1=1

4𝜋 𝜀0

𝑞1

|𝑟|2�̂�

2. The magnitude of the electric field:

3. The electric field in vector form:

Example Problem

|𝐸|= 14𝜋 𝜀0

𝑞|𝑟|2

¿ (9 ×1 09 N m2

C2 ) 2× 1 0− 9C(0.35 m ) 2¿147 N

C

𝐸=|𝐸|�̂�

𝐸𝑧

𝑥

𝑦

𝐸1=1

4𝜋 𝜀0

𝑞1

|𝑟|2�̂�

A penny carrying a small amount of positive charge Qp exerts an electric force F on a nickel carrying a large amount of positive charge Qn that is a distance d away (Qn > Qp ). Which one of the following is not true?

A. The electric force exerted on the penny by the nickel is also equal to F.B. The number of electrons in the penny is less than the number of protons in the penny.C. , if d is small compared to the size of the coins. D. , if d is large compared to the size of the

coins.

Clicker Question 1

A positive and a negative charge are separated by a distance r,

What are the directions of the forces on the charges?

Clicker Question 2

+q1 -q2

r

ChoiceA Left LeftB Right LeftC Left RightD Right Right

What is the magnitude of the self-force?

ChoiceA infiniteBC 0

The Coulomb Force

F

140

Q1Q2

r2 r̂0 = permittivity constant

221

041

rQQ

FF

• The force exerted by one point charge on another acts along line joining the charges.

• The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs.

Force on “2” by “1”

Force repulsive Force attractive

1

++2

r F21

+-2

r

F21

1

How Strong is the Coulomb Force

qFE /

[N/C]

tzyxEE ,,,

Electric Field

Electric field has units of Newton per Coulomb:

No ‘self-force’!

204

1rq

E

Point charge does not exert field on itself!

r 0, E

The net electric field at a location in space is a vector sum of the individual electric fields contributed by all charged particles located elsewhere.

The Superposition Principle

The electric field contributed by a charged particle is unaffected by the presence of other charged particles.

+q2

-q1

𝐸2𝐸1

𝐸𝑛𝑒𝑡

𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛𝑜𝑓 3+q3

The Superposition Principle+q2

𝐸2

𝐸3𝐸𝑛𝑒𝑡

-q1

+q3

𝐸 1 𝑃

The E of a Uniformly Charged Sphere

Can calculate using principle of superposition:

rrQEsphere ˆ

41

20

for r>R (outside)

0sphereE

for r<R (inside)

Recall this every night before bed!

Electric dipole:Two equally but oppositely charged point-like objects

What is the E field far from the dipole (r>s)?

+q-q

s

Example of electric dipole: HCl molecule

The Superposition PrincipleThe electric field of a dipole:

x

y

z+q-q

s

Choice of origin: use symmetry

Calculating Electric Field

Choice of the origin

1. E along the x-axis

xxx EEE ,,,1 1

40

qr s 2 2

1

40

qr s 2 2

E1,x 1

40

qr2 qrs qs2 / 4 qr2 qrs qs2 / 4r s 2 2 r s 2 2

E1,x 1

40

2qrsr s 2 2 r s 2 2

+q-q

s

r

𝐸1 ,𝑥

𝐸+ , 𝑥

𝐸− ,𝑥

220

,1

22

24

1

srsr

srqE x

if r>>s, then 222

22rsrsr

30

,12

41

rsqE x

0,0,24

13

01 r

sqE

While the electric field of a point charge is proportional to 1/r2, the electric field created by several charges may have a different distance dependence.

Approximation: Far from the Dipole

2. E along the y-axis

0,,2

0,0,2

0,,0

0,,2

0,0,2

0,,0

yssyr

yssyr

2

2

20,,

20,0,

20,,0

0,,2

0,0,2

0,,0

syrryssyr

yssyr

22

2

sy2

2

220

2

0,,2

24

1

sy

ys

sy

qE

22

220

2

0,,2

24

1

sy

ys

sy

qE

rrqE ˆ

41

20

+q-q s

𝐸2

𝐸−

y

𝐸+¿¿

𝑟+¿¿

2. E along the y-axis

22

220

2

0,,2

24

1

sy

ys

sy

qE

0,0,

2

41

23

22

02 s

sy

qEEE

0,0,1

2

41

23

22

02

sy

qsEEE

if r>>s, then

E2

140

qsr3 ,0,0

22

220

2

0,,2

24

1

sy

ys

sy

qE

at <0,r,0>

+q-q s

𝐸2

𝐸−

y

𝐸+¿¿

𝑟+¿¿

3. E along the z-axis

Due to the symmetry E along the z-axis must be the same as E along the y-axis!

0,0,4

13

02 r

qsE

at <0, r, 0>or <0, 0, r>

0,0,24

13

01 r

sqE

at <r, 0, 0>

y

xz

Other Locations

The Electric Field

rrqE ˆ

41

21

01

+ -

Point Charge:

Dipole: for r>>s :

E

140

2qsr3 ,0,0 at <r,0,0>

E

140

qsr3 ,0,0 at <0,r,0>

+q-qs

x

y

z at <0,0,r>

E

140

qsr3 ,0,0

Example ProblemA dipole is located at the origin, and is composed of particles with charges e and –e, separated by a distance 210-10 m along the x-axis. Calculate the magnitude of the E field at <0, 210-8, 0> m.

y

2Å

200Å

E=?

Since r>>s: 30

,1 41

rsqE x

E1,x 9 109 Nm2

C2

2 10 10 m 1.6 10 19 C

2 10 8 m 3

Using exact solution:

xCN4

,1 102.7 xE

CN4

,1 101999973.7 xE

Interaction of a Point Charge and a Dipole

0,0,24

13

0 dqsQEQF dipole

dipoleE

F

• Direction makes sense? - negative end of dipole is closer, so its net contribution is larger

• What is the force exerted on the dipole by the point charge? - Newton’s third law: equal but opposite sign

+q -q +Q

sF

Dipole Moment

0,0,24

13

01 r

qsE

0,0,4

13

02 r

qsE

x:

y, z:

Dipole moment: p = qs

qsp

0,0,24

13

01 r

pE

0,0,4

13

02 r

pE

, direction from –q to +q

r>>s

Dipole moment is a vector pointing from negative to positive charge

The electric field of a dipole is proportional to the

Dipole in a Uniform Field

EqF

Forces on +q and –q have the same magnitude but opposite direction

0 EqEqFnet

It would experience a torque about its center of mass.

Electric dipole can be used to measure the direction of electric field.

What is the equilibrium position?

Choice of SystemMultiparticle systems: Split into objects to include into system and objects to be considered as external.

To use field concept instead of Coulomb’s law we split the Universe into two parts:

• the charges that are the sources of the field

• the charge that is affected by that field

• Convenience: know E at some location – know the electric force on any charge:

Example: if E>3106 N/C air becomes conductor• Retardation

Nothing can move faster than light cc = 300,000 km/s = 30 cm/ns

• Can describe the electric properties of matter in terms of electric field – independent of how this field was produced.

Coulomb’s law is not completely correct – it does not contain time t nor speed of light c.

rrqE ˆ

41

20

rrqqF ˆ

41

221

0

v<<c !!!

F q

rE

A Fundamental Rationale