FUNDAMENTAL PHYSICS:
ELECTRICITY & MAGNETISM
A. Review:
1. Electric charges:
 The strength of a particle’s electrical interaction with objects around it depends on
its electric charge (q, unit: C), which can be either positive (+) or negative (-):
𝑆𝑎𝑚𝑒 𝑠𝑖𝑔𝑛 𝑐ℎ𝑎𝑟𝑔𝑒𝑠  𝑟𝑒𝑝𝑒𝑙. 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑔𝑛𝑠 𝑐ℎ𝑎𝑟𝑔𝑒𝑠  𝑎𝑡𝑡𝑟𝑎𝑐𝑡.
 Conductors = materials in which a significant number of charged particles are free
to move (free charges). The charged particles in nonconductors (insulators) are not
free to move (bound charges).
 Coulomb’s law: the force between 2 electric charges can be calculated:
(𝜀0 : permittivity constant)
1 𝑞1 𝑞2
𝑞1 𝑞2
𝐹=
=
𝑘
4𝜋𝜀0 𝑟 2
𝑟2
1 𝑞1 𝑞2
𝑉𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟𝑚: ⃗⃗⃗⃗⃗
𝐹12 =
𝑟̂
4𝜋𝜀0 𝑟 2 21
1
𝑚2
𝐶2
9
−12
𝑘=
≈ 8.99 ∗ 10 𝑁. 2 ; 𝜀0 = 8.85 ∗ 10
(
)
4𝜋𝜀0
𝐶
𝑁𝑚2
If multiple charges are presence: (Fi1 is force caused by particle number ith on
particle 1) note that this is a vector sum:
𝑛
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝐹1,𝑛𝑒𝑡 = ∑ ⃗⃗⃗⃗⃗
𝐹𝑖1
𝑖=2
 Quantization of charge: The charge of a particle can always be written as ne.
𝑞 = 𝑛𝑒 (𝑛 = ±1, ±2, … )
𝑒 = 1.602 ∗ 10−19 𝐶
 Conservation of charge: The net electric charge of any isolated system is always
conserved.
2. Electric fields:
 Definition: The electric field at any point is defined in terms of the electrostatic
force that would be exerted on a positive test charge q0 placed there:
𝐸⃗ =
𝐹
𝑞0
 Electric field lines: visualizing the direction and magnitude of electric fields: The
electric field vector at any point is tangent to a field line through that point. The
density of field lines in any region is proportional to the magnitude of the electric
field in that region.
 Field due to a point charge: Magnitude of E due to a point charge q at a distant r
from it is: (direction: points away if q>0; points towards if q<0)
1 |𝑞|
𝐸=
2
4𝜋𝜀0 𝑟
 Field due to a continuous distribution of charge: we treat small element dq as point
charges, and take the integral over the whole distribution to obtain the result:
|𝑑𝑞|
1
∫ 2
𝐸=
4𝜋𝜀0
𝑟
 Field due to a discrete distribution of charge: each charge qi contribute Ei to the
total E (vector):
𝑛
⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗
𝐸
𝑛𝑒𝑡 = ∑ 𝐸𝑖
𝑖=1
 Force on q charge in E field: 𝐹 = 𝑞𝐸⃗
 Dipole: consists of two particles with charges of equal magnitude q but opposite
sign, separated by a small distance d.
o Dipole moment: 𝑝 = 𝑞𝑑 in magnitude, direction points from positive to
negative
o E field due to dipole: (at distance z, on dipole axis)
𝐸=
1 𝑝
2𝜋𝜀0 𝑧3
o Dipole in E field: the field exert torque on the dipole: 𝜏 = 𝑝 × 𝐸⃗ , potential
energy associated: 𝑈 = −𝑝. 𝐸⃗ (least when dipole align with E and greatest
when E opposite)
 E field of a uniformly charged ring: (at point on central axis, distant z from the
ring)
𝑞𝑧
𝐸=
3
4𝜋𝜀0 (𝑧 2 + 𝑅2 )2
 E field of a uniformly charged disk: (at point on central axis, at distant z from the
disk)
𝜎
𝑧
𝐸=
(1 −
)
2𝜀0
√𝑧 2 + 𝑅2
3. Gauss’ law:
 Electric flux: The flow of E through a perpendicular surface A in a unit of time.
o For an uniform E field flowing through a flat surface A, the electric flux is
calculated:
Φ = 𝐸⃗ . 𝐴
o For the case where E is not uniform and the surface A is not flat, we will have
to divide the surface A into small elements dA. On the small surface dA, the
variation in E is small enough so we can consider it to be uniform on dA, so
the flux through the surface dA is:
𝑑Φ = 𝐸⃗ . 𝑑𝐴
Then we take the integral over the area to sum up all dA, so that we obtain the
result:
Φ = ∫ 𝐸⃗ . 𝑑𝐴
 Gauss’ law:
o To apply Gauss’ law we need a Gaussian surface: A closed surface, enclosing a
charge q.
o Gauss’ law: (the circle integral symbol means integrating around the surface)
𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
Φ=
= ∮ 𝐸⃗ . 𝑑𝐴
𝜀0
(𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 : 𝑐ℎ𝑎𝑟𝑔𝑒 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
 Applications of Gauss’ law: (All of this can be proven using Gauss’ law easily,
proofs are written in the books. If you don’t know how to, at least try to memorize
them):
o Excess charge on an isolated conductor is located entirely on the surface of it
o The external electric field near the surface of a charged conductor is
perpendicular to the surface and has magnitude:
𝐸𝑜𝑢𝑡 =
𝜎
𝜀0
Internal electric field of a conductor is zero:
𝐸𝑖𝑛 = 0
o E field at a point (with perpendicular distance r from the line) due to an infinite
line of charge (linear charge density) is perpendicular to the line and has
magnitude:
𝐸=
λ
2𝜋𝜀0 𝑟
o E field due to infinite non-conducting sheet with uniform charge density is
perpendicular to the sheet and has magnitude:
𝐸=
𝜎
2𝜀0
o E outside spherical shell of charge q, radius R, at distance r from the center is
radial and has magnitude
𝐸=
1 𝑞
4𝜋𝜀0 𝑟2
E inside a spherical shell: 𝐸 = 0
o E field inside uniform sphere of charge (radius r, r < R) is radial and has
magnitude:
𝐸=
1 𝑞
𝑟
4𝜋𝜀0 𝑅3
B. Exercises:
(from Halliday’s book)
1. Chapter 21: Coulomb’s Law
𝐹=
1 𝑞1 𝑞2
4𝜋𝜀0 𝐹
→
𝑟
=
√
≈ 1.39 (𝑚)
4𝜋𝜀0 𝑟 2
𝑞1 𝑞2
a) Looking at the graph, as the position of charge 3 moves closer to 0 (position of
charge 1), the force tends to infinite  there’s a repulsive force between charge 1
and 3 => charge 1 is positive
b) The point where F 3,net =0 is located in between charge 1 and 2, knowing that
charge 1 and 3 is positive, the only scenario here is charge 2 is also positive.
We know that the magnitude:
𝐹3,𝑛𝑒𝑡 = |𝐹13 − 𝐹23 |
Substituting value at which F3net=0 we can obtain:
1 𝑞1 𝑞3
1 𝑞2 𝑞3
=
4𝜋𝜀0 (𝑥𝑠 )2 4𝜋𝜀0 (3𝑥𝑠 )2
4
4
Solving this we have the answer:
𝑞2
=9
𝑞1
The force on charge q3 is as shown in the figure. From that we can determine the net
force on each component:
1 |𝑞2 𝑞3 |
1 |𝑞4 𝑞3 |
𝐹3𝑥 = 𝐹23𝑥 + 𝐹43 =
𝑠𝑖𝑛45° +
= 0.17𝑁
4𝜋𝜀0 (𝑎√2)2
4𝜋𝜀0 𝑎2
1 |𝑞2 𝑞3 |
1 |𝑞1 𝑞3 |
𝐹3𝑦 = 𝐹23𝑦 − 𝐹13 =
𝑐𝑜𝑠45° −
= −0.0465𝑁
4𝜋𝜀0 (𝑎√2)2
4𝜋𝜀0 𝑎2
{
a) 𝐹 =
1
𝑞2
4𝜋𝜀0 𝑟 2
= 8.99 × 10−19 𝑁
b) 𝑞 = 𝑛𝑒 → 𝑛 = 𝑞 ⁄𝑒 = 625 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝑞 = 𝑛𝑒 𝑤𝑖𝑡ℎ 𝑛 = 1500 (𝑚−2 𝑠 −1 )
𝑑𝑞
𝑖𝑜𝑛 1 𝑚2 =
=𝑞
𝑑𝑡
→ 𝐼𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 𝑖𝑜𝑛 1 𝑚2 × 𝑆𝑒𝑎𝑟𝑡ℎ =
a) The rod is horizontal and balanced: Total bending moment on the rod should be 0:
𝐿
𝐿
𝐿
𝐹1 ∗ + 𝑊 ∗ (𝑥 − ) = 𝐹2 ∗
2
2
2
1 𝑞𝑄 𝐿
𝐿
1 2𝑞𝑄 𝐿
↔
+ 𝑊 (𝑥 − ) =
4𝜋𝜀0 ℎ2 2
2
4𝜋𝜀0 ℎ2 2
𝐿
𝑞𝑄𝐿
↔𝑥= +
2 8𝜋𝜀0 ℎ2 𝑊
b) Balance of the vertical force:
𝐹1 + 𝐹2 = 𝑊
↔
1 𝑞𝑄
1 2𝑞𝑄
1 3𝑞𝑄
+
=𝑊 ↔ℎ=√
2
2
4𝜋𝜀0 ℎ
4𝜋𝜀0 ℎ
4𝜋𝜀0 𝑊
1 𝑒2
𝑚𝑒2
𝐹𝑒 =
; 𝐹 =𝐺 2
4𝜋𝜀0 𝑟 2 𝑔
𝑟
2
𝐹𝑔 4𝜋𝜀0 𝐺𝑚𝑒
→ =
𝐹𝑒
𝑒2
2. Chapter 22: Electric Field
(Note: E2 and E3, E1 and E4 have same direction and magnitude so I can’t distinguish
them in the graph)
Looking at the graph:
𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛: 𝐸𝑛𝑒𝑡𝑥 = 𝐸1 𝑠𝑖𝑛45 + 𝐸2 𝑠𝑖𝑛45 + 𝐸3 𝑠𝑖𝑛45 + 𝐸4 𝑠𝑖𝑛45
{
𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛: 𝐸𝑛𝑒𝑡𝑦 = −𝐸1 𝑠𝑖𝑛45 + 𝐸2 𝑠𝑖𝑛45 + 𝐸3 𝑠𝑖𝑛45 − 𝐸4 𝑠𝑖𝑛45
1 𝑞2
1 𝑞3
1 𝑞4
√2 1 𝑞1
(
+
+
+
)
2 4𝜋𝜀0 𝑎2 4𝜋𝜀0 𝑎2 4𝜋𝜀0 𝑎2 4𝜋𝜀0 𝑎2
2
2
2
2
1 𝑞2
1 𝑞3
1 𝑞4
√2 −1 𝑞1
𝐸𝑛𝑒𝑡𝑦 =
(
+
+
−
)
2 4𝜋𝜀0 𝑎2 4𝜋𝜀0 𝑎2 4𝜋𝜀0 𝑎2 4𝜋𝜀0 𝑎2
{
2
2
2
2
𝐸𝑛𝑒𝑡𝑥 =
Direction: perpendicular to x axis, points in negative y.
Magnitude: (theta is the angle between E+ and E)
𝑑
1
𝑞
1
𝑞𝑑
2
𝐸 = 2𝐸+ 𝑐𝑜𝑠𝜃 = 2
=
2
2
3
2
4𝜋𝜀0 𝑑
4𝜋𝜀0 𝑑
+ 𝑟 2 √𝑑 + 𝑟 2
( + 𝑟 2 )2
4
4
4
2
𝑠𝑖𝑛𝑐𝑒 𝑟 ≫ 𝑑, 𝑤𝑒 𝑐𝑎𝑛 𝑎𝑝𝑝𝑟𝑜𝑥 (
→𝐸=
3
2
𝑑
+ 𝑟2) ≈ 𝑟3
4
1 𝑞𝑑
4𝜋𝜀0 𝑟 3
a) At z=0, we can see that the field cancel out due to symmetry  E=0
b) At z=∞, using the formula for E for a charged ring:
𝑞𝑧
𝐸=
3
4𝜋𝜀0 (𝑧 2 + 𝑅 2 )2
As z increases, the denominator increases much faster => the whole fraction tends
to 0
c) E max when
𝑑𝐸
𝑅
=0↔𝑧=
𝑑𝑧
√2
d) Substitute the values in
𝑄𝑅
𝑁
7
𝐸=
=
4.33
×
10
(
)
2
3
𝐶
𝑅
2
4√2𝜋𝜀0 ( + 𝑅 )2
2
Since the charge is uniformly distributed, we just need to calculate the surface area
ratio between the ring and the disk
𝑞=𝑄
𝐴𝑟𝑖𝑛𝑔
2𝜋𝑟𝑤
=𝑄
𝐴𝑑𝑖𝑠𝑘
𝜋𝑅2
When the electron enter the region, the electric field acts on it a retarding force of
magnitude: 𝐹 = 𝑒𝐸.
a) 2nd law of Newton: 𝐹 = 𝑚𝑎 = 𝑒𝐸 => 𝑎 =
𝑒𝐸
𝑚
Electron stop momentarily: v=0
𝑣0 2
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑: 𝑠 =
2𝑎
→{
𝑣0
𝑇𝑖𝑚𝑒 𝑒𝑙𝑎𝑝𝑠𝑒𝑑: 𝑡 =
𝑎
b) The region is 8mm long (d=8), using the kinetic work theorem, we get
|∆𝐾| = |𝐹. 𝑑|
|∆𝐾| 2|𝑒|𝐸𝑑
=> 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑙𝑜𝑠𝑡:
=
𝐾
𝑚𝑒 𝑣02
a) Direction of F net due to bee is towards the bee
𝑄
𝑞−
𝑞+
−10
( 2−
𝑁
2 )| = 2.562 × 10
4𝜋𝜀0 𝐷1
𝐷1
( + 𝐷2 )
4
2
b) Direction of F net due to the stigma is towards the stigma:
𝐹𝑛𝑒𝑡,𝑏𝑒𝑒 = |𝐹− − 𝐹+ | = |
𝑞𝑠 𝑞+
𝑞−
𝐹𝑛𝑒𝑡,𝑠𝑡𝑖𝑔𝑚𝑎 = |𝐹+ − 𝐹− | = |
( 2 −
)| = 3.056 × 10−8 𝑁
(𝑙 + 𝐷2 )2
4𝜋𝜀0 𝑙
c) 𝐹𝑛𝑒𝑡,𝑠𝑡𝑖𝑔𝑚𝑎 > 𝐹𝑛𝑒𝑡,𝑏𝑒𝑒 => 𝑇ℎ𝑒 𝑔𝑟𝑎𝑖𝑛 𝑤𝑖𝑙𝑙 𝑙𝑒𝑎𝑣𝑒 𝑡ℎ𝑒 𝑏𝑒𝑒
Dipole moment: 𝑝 = 𝑞𝑑
Potential energy
𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝐸: 𝑈−= −𝑝𝐸
𝐴𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝐸: 𝑈+= 𝑝𝐸
=> 𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒: ∆𝑈 = 2𝑝𝐸 = 2𝑞𝑑𝐸
{
The electric field at a point on the axis of a uniformly charged ring,
a distance z from the ring center, is:
𝑞𝑧
𝐸=
3
4𝜋𝜀0 (𝑧 2 + 𝑅2 )2
𝑞𝑧
𝑆𝑖𝑛𝑐𝑒 𝑅 ≫ 𝑧 => 𝑧 2 + 𝑅2 → 𝑅2 ↔ 𝐸 =
4𝜋𝜀0 𝑅3
The E field will act a force on the eletron, the force is:
−𝑒𝑞𝑧
𝐹=
4𝜋𝜀0 𝑅3
This F force is a resorting force, we can imagine this situation like a spring oscillation,
where F=-kx. Here zx and the rest k
𝜔=√
𝑘
𝑒𝑞
=√
𝑚
4𝜋𝜀0 𝑚𝑅3
3. Chapter 23: Gauss’ Law
a) Top face: 𝑦 = 2 => 𝐸⃗ = 4𝑖 − 18𝑗 uniform through top. 𝑑𝐴 = 𝑑𝐴. 𝑗
𝑁𝑚2
Φtop = ∫ 𝐸⃗ . 𝑑𝐴 = 𝐸⃗ . 𝐴 = (4𝑖 − 18𝑗)𝐴. 𝑗 = −18 × 2 = −72 (
)
𝐶
b) Bottom face: 𝑦 = 0 => 𝐸⃗ = 4𝑖 − 6𝑗 uniform through bottom. 𝑑𝐴 = 𝑑𝐴. (−𝑗)
𝑁𝑚2
2
⃗
⃗
Φbottom = ∫ 𝐸 . 𝑑𝐴 = 𝐸 . 𝐴 = (4𝑖 − 6𝑗)𝐴. −𝑗 = 6 × 2 = 24 (
)
𝐶
c) Left face: 𝑑𝐴 = 𝑑𝐴. (−𝑖)
𝑁𝑚2
2
⃗
⃗
Φ𝑙𝑒𝑓𝑡 = ∫ 𝐸 . 𝑑𝐴 = 𝐸 . 𝐴 = (4𝑖 − 𝐸𝑦𝑗)𝐴. −𝑖 = −4 × 2 = −16 (
)
𝐶
d) Backface: E has no z component => No E through back face=> Flux=0
2
e) Φ = ∑ Φeach = −72 + 24 + 16 − 16 = −48 (
Φ𝑛𝑒𝑡 =
𝑞
𝜀0
𝑁𝑚 2
𝐶
)
At x=2cm, shell 1 have no contribution towards E net, we just have to calculate for
shell 2. (direction is –x because shell 2 is positively charged -> field radial out)
𝑞2
𝜎2 𝑅22
𝑁
4
⃗𝐸 = −
𝑖̂
=
−
𝑖̂
=
−2.8
×
10
(
) 𝑖̂
4𝜋𝜀0 (𝐿 − 2)2
𝜀0 (𝐿 − 2)2
𝐶
Φ=
a) 𝑞𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝜀0 Φ1
b) 𝑞𝑠ℎ𝑒𝑙𝑙 𝐴 = 𝜀0 Φ2
c) 𝑞𝑠ℎ𝑒𝑙𝑙 𝐵 = 𝜀0 Φ3
𝑞
𝜀0
Surface density
𝑄
𝑄
=
𝐴 𝜋𝐷 2
E field just outside the surface of a spherical shell:
𝜎=
𝐸=
1 𝑄
𝜋𝜀0 𝐷2
Using Gauss law we can proof that the E field (outside) of these shells have formula:
λ
𝐸=
2πε0 𝑟
E inside of a shell is still zero
In the first region of the graph, E is only the contribution of the inner shell
Using this we can solve for lambda 1
λ1
𝐸1 =
2𝜋𝜀0 𝑟
𝐶
Sub in the value r=3.5, E=10^3 => λ1 = 1.95 × 10−9 ( )
𝑚
E outside is the sum of E cylinder and cylindrical shell
λ1 + λ2
𝐸 = 𝐸1 + 𝐸2 =
2𝜋𝜀0 𝑟
𝐶
Sub in the value r=3.5, E= -2.10^3 => λ2 = −5.83 × 10−9 ( )
𝑚
Using Gauss law we can proof that the E field (outside) of these shells have formula:
λ
𝐸=
2πε0 𝑟
E inside of a shell is still zero
a) 𝑟 = 4𝑐𝑚 < 𝑟2 =>Shell 2 don’t contribute E in this case
λ1
𝑁
=> 𝐸 =
= 2.29 × 106 ( ) > 0 => 𝑅𝑎𝑑𝑖𝑎𝑙 𝑜𝑢𝑡𝑤𝑎𝑟𝑑
2πε0 𝑟
𝐶
b) r = 8 cm > r_2 => Both shell contribute:
λ1 + λ2
𝑁
=> 𝐸 = 𝐸1 + 𝐸2 =
= −4.5 × 105 ( ) < 0 => 𝑅𝑎𝑑𝑖𝑎𝑙 𝑖𝑛𝑤𝑎𝑟𝑑
2πε0 𝑟
𝐶
This is equivalent to 1 infinite sheet of density sigma (denote as 1) + 1 disk (size=hole)
of density (-) sigma (denote as 2).
𝜎
𝜎
𝑧
𝜎𝑧
𝑁
𝐸⃗ = 𝐸⃗1 + 𝐸⃗2 = 𝑧̂ (
+ (−
) (1 −
)) = 𝑧̂
= 0.208𝑧̂ ( )
2𝜀0
2𝜀0
𝐶
2𝜀0 √𝑧 2 + 𝑅2
√𝑧 2 + 𝑅2
The first part of the graph is the Electric field of the particle (Because there’s a
discontinuous region between the two part, we can conclude that it is caused by going
through a shell). That means with x<2.5 the value of E is only dependent on the
charged particle:
𝑞
2.4 2
𝐸𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 =
= ( ) 𝐸2.4
4𝜋𝜀0 𝑟 2
𝑟
(We use the point x=2.4 and E=2.10^7 to calculate E of charged particle)
The second part, electric field is the contribution of both the particle and the shell
2.4 2
𝐸 = 𝐸𝑠ℎ𝑒𝑙𝑙 + 𝐸𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝐸𝑠ℎ𝑒𝑙𝑙 + ( ) 𝐸2.4
𝑟
Using the point x=3, E=8.10^7, we can calculate 𝐸𝑠ℎ𝑒𝑙𝑙 = 6.6 × 107 at x=3.
Finally, to calculate Q on shell
2
𝑄 = 4𝜋𝜀0 𝑟𝑠ℎ𝑒𝑙𝑙
𝐸𝑠ℎ𝑒𝑙𝑙 = 6.6 × 106 𝐶
Since the charge is uniformly distributed, the amount of charge in any radius r is Q
times the ratio of the volume r and total sphere.
𝑉𝑟
𝑟3 𝑄
𝑞=𝑄
=𝑄 3=
𝑉
𝑅
8
E field on the surface of sphere:
1 𝑄
𝐸=
4𝜋𝜀0 𝑅2
E field at radius r:
1 𝑄
𝑟
4𝜋𝜀0 𝑅 3
𝐸𝑟 𝑟 1
𝑅𝑎𝑡𝑖𝑜:
= =
𝐸
𝑅 2
𝐸𝑟 =