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Things you must know
Relationship between electric field and electric force Conservation of charge
Electric field of a point charge The Superposition Principle
Relationship between magnetic field and magnetic force
Magnetic field of a moving point charge
Other Fundamental Concepts
~a =
d~v
dt
d~p
dt
Fnet and
d~p
dt
≈ m~a if v << c
∆Uel = q∆V ∆V = −
f i
E~ • d~l ≈ −
(Ex∆x + Ey∆y + Ez ∆z)
Φel =
E • ˆndA Φmag =
B • ˆndA
∮ ~ E • ndAˆ =
qinside
B • ˆndA = 0
|emf| =
EN C • d
l =
dΦmag
dt
B • d
l = μ 0
Iinside path
B • d
l = μ 0
[
Iinside path + 0
d
dt
E • ndAˆ
]
Specific Results
Edipole,axis
4 π 0
2 qs
r 3
(on axis, r s)
Edipole,⊥
4 π 0
qs
r 3
(on ⊥ axis, r s)
Erod
4 π 0
Q
r
r
2
2
(r ⊥ from center) electric dipole moment p = qs, ~p = α
Eapplied
E~
rod
∣ ≈^
4 π 0
2 Q/L
r
(if r L)
E~
ring
∣ =^
4 π 0
qz
(z 2
(z along axis)
Edisk
Q/A
[
|z|
(z 2
] ∣
Edisk
Q/A
[
|z|
R
]
Q/A
(if z R)
Ecapacitor
∣ ≈^
Q/A
(+Q and −Q disks)
Ef ringe
∣ ≈^
Q/A
s
2 R
just outside capacitor
∆ B~ =
μ 0
4 π
I∆
` × ˆr
r 2
(short wire) ∆ F~ = I∆
l × B~
Bwire
μ 0
4 π
LI
r
r
2
2
μ 0
4 π
2 I
r
(r L)
Bwire
Bearth
∣ tan θ
B~
loop
∣ =^
μ 0
4 π
2 IπR
2
(z 2
μ 0
4 π
2 IπR
2
z 3
(on axis, z R) μ = IA = IπR
2
B~
dipole,axis
∣ ≈^
μ 0
4 π
2 μ
r 3
(on axis, r s)
B~
dipole,⊥
∣ ≈^
μ 0
4 π
μ
r 3
(on ⊥ axis, r s)
Erad =
4 π 0
−q~a⊥
c 2 r
ˆv =
Erad ×
Brad
Brad
Erad
c
i = nA¯v I = |q| nA¯v v¯ = uE
σ = |q| nu J =
I
A
= σE R =
L
σA
Edielectric =
Eapplied
K
∆V =
q
4 π 0
[
rf
ri
]
due to a point charge
I =
|∆V |
R
for an ohmic resistor (R independent of ∆V ); power = I∆V
Q = C |∆V | K ≈
1 2
mv
2 if v c
circular motion:
d~p
dt ⊥
|~v|
R
|~p| ≈
mv
2
R
Math Help
~a ×
b = 〈ax, ay, az 〉 × 〈bx, by, bz 〉
= (ay bz − az by)ˆx − (ax bz − az bx)ˆy + (ax by − ay bx)ˆz
dx
x + a
= ln (a + x) + c
dx
(x + a) 2
a + x
dx
(a + x) 3
2(a + x) 2
a dx = ax + c
ax dx =
a
x
2
ax
2 dx =
a
x
3
xdx
(x
2
2 )
3 / 2
(x
2
2 )
1 / 2
a
x(x + b)
dx =
a
b
x
x + b
) dx
Constant Symbol Approximate Value
Speed of light c 3 × 10
8 m/s
Gravitational constant G 6. 7 × 10
− 11 N · m
2 /kg
2
Approx. grav field near Earth’s surface g 9 .8 N/kg
Electron mass me 9 × 10
− 31 kg
Proton mass mp 1. 7 × 10
− 27 kg
Neutron mass mn 1. 7 × 10
− 27 kg
Electric constant
4 π 0
9 × 10
9 N · m
2 /C
2
Epsilon-zero 0 8. 85 × 10
− 12 (N · m
2 /C
2 )
− 1
Magnetic constant
μ 0
4 π
1 × 10
− 7 T · m/A
Mu-zero μ 0 4 π × 10
− 7 T · m/A
Proton charge e 1. 6 × 10
− 19 C
Electron volt 1 eV 1. 6 × 10
− 19 J
Avogadro’s number NA 6. 02 × 10
23 molecules/mole
Atomic radius Ra ≈ 1 × 10
− 10 m
Proton radius Rp ≈ 1 × 10
− 15 m
E to ionize air Eionize ≈ 3 × 10
6 V/m
BEarth (horizontal component) BEarth ≈ 2 × 10 − 5 T
Problem 1
- [10 pts] An electron with charge −e is placed at position 〈−a, b, 0 〉, and a proton with charge +e is placed
at 〈a, b, 0 〉. Determine where to place a point charge of 3e so that the electric field at the origin 〈 0 , 0 , 0 〉
vanishes.
- [20 pts] In the limit where a b the electron proton pair can be treated like a dipole. Using this dipole
approximation for the electron-proton pair (and the Coulomb law for the 3e point charge), determine the
net electric field (from the dipole and the 3e point charge) at location 〈 0 , 2 b, 0 〉.
[10 pts] A second, identical rod is placed on top of the first forming a “T” as indicated in the diagram.
Calculate the new net electric field at the observation location. You can ignore the thickness of the rods.
Problem 3
A small glass ball acquires a uniform negative charge −Q. A distance d to the right of the glass ball is a permanent
dipole, with dipole moment |Qs|. Both the ball and the dipole can not move and the distance between the objects
is large d s. If any of the electric fields you are asked to draw is zero, state this explicitly.
- [12 pts] At the locations A, B, C (marked “x”) draw and label two arrows, representing the electric field due
to the ball and the the field from the dipole. The relative lengths of these arrows should correspond to the
relative magnitude of the fields between the different locations.
- [6 pts] A neutral metal block is placed between the two objects. Draw the approximate charge distribution
for the metal block on the diagram below.
- [8 pts] At the center of the metal block draw and label four arrows (The relative lengths of these arrows
should correspond to the relative magnitude of the fields):
The electric field due to the permanent dipole
E (^) dipole
The electric field due to the glass ball
E (^) ball
The electric field due to the charges in and/or on the metal block
E (^) metal
The net electric field
E (^) net
- [4 pts] Calculate the magnitude of the electric field at the center of the block due to the charge in and/or
on the metal block.
