1. Example Problem Solution:
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?
Solution:
Distance and direction:
𝐸 1 = 1 4𝜋 𝜀 0 𝑞 1 𝑟 2 𝑟
𝑟 = 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑_𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 − 𝑠𝑜𝑢𝑟𝑐𝑒_𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛
𝑟 = −0.2, −0.2, −0.2 − 0, 0, 0
Note: r – direction is from source to point of observation.
𝑟 = (−0.2 ) 2 +(−0.2 ) 2 +(−0.2 ) 2 =0.35m
𝑟 = 𝑟 𝑟 = −0.2, −0.2, − = −0.57,−0.57,−0.57

2. Example Problem 2. The magnitude of the electric field:
𝐸 1 = 1 4𝜋 𝜀 0 𝑞 1 𝑟 2 𝑟
𝐸 = 1 4𝜋 𝜀 0 𝑞 𝑟 2
= 9× N m 2 C ×1 0 −9 C m 2
=147 N C
3. The electric field in vector form: 𝐸 𝑧 𝑥 𝑦
=147 N C −0.57,−0.57,−0.57
𝐸 = 𝐸 𝑟
𝐸 = −84,−84,−84 N C

3. Clicker Question 1
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?
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. 𝑭≈ 𝟏 𝟒𝝅 𝜺 𝟎 𝑸 𝒑 𝑸 𝒏 𝒅𝟐 , if d is small compared to the size of the
coins.
D. 𝑭≈ 𝟏 𝟒𝝅 𝜺 𝟎 𝑸 𝒑 𝑸 𝒏 𝒅𝟐 , if d is large compared to the size of the
Answer: C

4. Clicker Question 2 r +q1 -q2 Choice 𝑭 𝒐𝒏 𝒒 𝟏 𝑭 𝒐𝒏 𝒒 𝟐 A Left B Right C
A positive and a negative charge are separated by a distance r,
What are the directions of the
forces on the charges? r
+q1
-q2
What is the magnitude of
the self-force?
Choice
𝑭 𝒐𝒏 𝒒 𝟏
𝑭 𝒐𝒏 𝒒 𝟐 A
Left B
Right C D
Choice
𝑭 𝒐𝒏 𝒒 𝟐
𝒅𝒖𝒆 𝒕𝒐 𝒒 𝟐 A
infinite B
k 𝒒 𝟏 𝒒 𝟐 /𝒓𝟐 C
Answer: D

5. The Coulomb Force e0 = permittivity constant Force repulsive
Force attractive 1 + 2 r
F21 -
Force on “2” by “1”
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.

6. How Strong is the Coulomb Force

7. Electric Field
[N/C]
Electric field has units of Newton per Coulomb:

8. No ‘self-force’! Point charge does not exert field on itself!
Physically makes sense – it cannot start itself moving, there is no way to decide in which way to move.

9. The Superposition Principle
𝐸 2
𝐸 1
𝐸 𝑛𝑒𝑡
𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 3
+q2
+q3
-q1
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 electric field contributed by a charged particle is unaffected by the presence of other charged particles.

10. The Superposition Principle
+q2
-q1
+q3
𝐸 1 𝑃
𝐸 𝑛𝑒𝑡
𝐸 3
𝐸 2

11. The E of a Uniformly Charged Sphere
Can calculate using principle of superposition:
for r>R (outside)
Uniformly charged sphere acts like a point charge! Not only its field is the same (outside), but it also interacts with other charge particles as a point charge.
for r<R (inside)
Recall this every night before bed!

12. The Superposition Principle
The electric field of a dipole:
Electric dipole:
Two equally but oppositely charged point-like objects +q -q s
Example of electric dipole: HCl molecule
What is the E field far from the dipole (r>s)?

13. Calculating Electric Field
Choice of the origin x y z +q -q s
Choice of origin: use symmetry

14. 1. E along the x-axis +q -q s
𝐸 1,𝑥
𝐸 −,𝑥
𝐸 +,𝑥 r

15. Approximation: Far from the Dipole
if r>>s, then
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.

16. 2. E along the y-axis
𝐸 +
𝐸 2
𝑟 +
𝐸 − y -q +q s

17. 2. E along the y-axis 𝐸 + 𝐸 2 𝑟 + 𝐸 − y -q +q s if r>>s, then
𝐸 +
𝐸 2
𝑟 +
𝐸 − y
Ask now about coordinate <0,0,r> - make them think about symmetry.
if r>>s, then
at <0,r,0> -q +q s

18. 3. E along the z-axis y z x at <r, 0, 0> at <0, r, 0>
or <0, 0, r> z x
Point out difference – direction and magnitude
Due to the symmetry E along the z-axis must be the same as E along the y-axis!

19. Other Locations
Ask students where is + and where is -.

20. The Electric Field y x z Point Charge: Dipole: s -q +q + -
for r>>s :
at <r,0,0>
at <0,r,0> +q -q s x y z
at <0,0,r>
Note that the magnitude of the electric field at a location along the y axis is HALF of the magnitude of the electric field at a location the same distance away on the x axis!

21. Example Problem y
A dipole is located at the origin, and is composed of particles with charges e and –e, separated by a distance 210-10 m along the x-axis. Calculate the magnitude of the E field at <0, 210-8, 0> m.
E=?
Since r>>s:
200Å x
A dipole is located at the origin, and is composed of particles with charge e and –e, separated by a distance 210-10 m along the x-axis. Calculate the magnitude of the E field at <0,210-8,0> m. 2Å
Using exact solution:

22. Interaction of a Point Charge and a Dipole+q -q +Q s
𝑑≫𝑠
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
Why is it possible? Total charge of dipole is zero! – the key lays in distance dependence

23. Dipole Moment x: r>>s y, z:
The electric field of a dipole is proportional to the
Dipole moment: p = qs
Note – aproximate
, direction from –q to +q
Dipole moment is a vector pointing from negative to positive charge

24. Dipole in a Uniform Field
Forces on +q and –q have the same magnitude but opposite direction
It would experience a torque about its center of mass.
What would be equi
What is the equilibrium position?
Electric dipole can be used to measure the direction of electric field.

25. Choice of System
Multiparticle 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

26. A Fundamental Rationale
Convenience: know E at some location – know the electric force on any charge:
Can describe the electric properties of matter in terms of electric field – independent of how this field was produced.
Example: if E>3106 N/C air becomes conductor
Retardation
Nothing can move faster than light c
c = 300,000 km/s = 30 cm/ns
Coulomb’s law is not completely correct – it does not contain time t nor speed of light c.
Example: Suppose I am negatively charged sphere, move to one side ask student in back row to show the direction of E due to this charge.
Then move to the other side at ~speed of light and ask student what will happen to E. It is ~5 meters, so it takes 15 ns for E to change direction after I moved!
More drastic example – pretend that I hold electron and positron (dipole), ask student to show E. Collapse (annihilate) charges and count time.
Conclusion: E can exist independently of charges!!!
Does not contain v – works only when speed is << c
v<<c !!!