ELECTRIC FIELDS For x ≫ a, a can be neglected: Uniformly charged disk With m = ρV and d = ρ1 V , we have:

Coulomb’s law 1 Q ϵρ1

⃗E = ı̂ ρ=
⃗F12 = ke q1 q2 r̂12 4πϵ0 x2 ϵ−1
ϵr 2 ρ1 is the density of the dielectric mate-
1 rial and Phd is the apparent weight of the
where ke = = 9 · 109 m2 /C2 , ϵ0 = object.
4πϵ0
8.854 × 10−12 GAUSS’S LAW
Electric field vector For a ≫ x, x2 /a2 can be neglected: "
x
#
Electric flux
Ex = 2ke πσ 1 − =E
Due to point charges ⃗E = λ (R2 + x2 )1/2 Through surface area A !
ı̂ qσ
2πϵ0 x α = arctan
⃗ Electric dipole in an electric field ΦE = EA cos θ
⃗E = ke q ⃗r = F 2ϵϵ0 mg
r2 q Circular rod θ: angle between E and the normal of A,
measured in radian.
Two oppositely charged conducting
with r
⃗ is a unit vector directed from the plates with surface charge densi-
A: surface area
charge towards the point in question. ties +σ and −σ
Through a closed Gaussian surface
I I
NOTE: The electric field is directed radi-
ΦE = E cos θ dA = E⊥ dA
ally outward from a positive charge and
radially inward towards a negative charge. A plastic rod with uniform charge −Q I
⃗ = Qencl
Due to continuous charge distribu- is bent in a 2θ circular arc of radius r = ⃗E · d A
ϵϵ0
and symmetrically paced across an x axis The electric dipole moment is the vec-
tions with the origin at the center of curvature
Z tor directed from −q to +q along the line Electric field of various symmetric
⃗E = ke dq r̂ P of the rod. joining the charges and having magni- charge distributions
r 2 tude:
p ≡ 2aq
Infinite sheet of charge with uni-
Electric fields of various special form charge per unit area σ
cases Torque on an electric dipole in an exter-
Due to symmetry, the y-components of nal electric field: σ
Charged rod along the x-axis E=
all segments cancel each other out. τ⃗ = ⃗p × ⃗E ϵϵ0
k|Q| sin θ ⃗ = kQ sin θ î Potential energy of the system of an elec- Solid insulating sphere with ra-
E= E
θr 2 θr 2 tric dipole in an external electric field: dius R and charge Q uniformly dis-
tributed throughout volume
Uniformly charged disk of radius R UE = −⃗p · ⃗E
with a concentric hole of radius r
Two identical spheres in a dielectric P can be any point on the sheet.
ke Q material σ
E= Find the density of each spheres so that E=
a (l + a) 2ϵϵ0
they have equal deflection angles.
Charged rod along the y-axis, cen- AA′ is an infinite sheet of sur-
tered at the origin face charge density σ . B is an
σ m-gram sphere carrying a similar
E= q
2 charge (similar sign) of q with the
2ϵ0 ϵ 1 + r 2
R sheet. Find the deflection angle α.
Uniform ring of charge Point P is at distance r from the center
of the sphere of radius R.

For r < R:
When placed in air: ρr Qr
E= =
F q2 3ϵϵ0 4πϵϵ0 R3
tan α = =
P 4πϵ0 r 2 mg For r > R:
ke x
Ex =  3/2 Q = E When placed in a dielectric material: Q
1 Q q2 E=
⃗E = √ ı̂ a2 + x 2 F′ 4πϵϵ0 r 2
4πϵ0 x x2 + a2 tan α = = 2
Phd 4πϵϵ0 r (mg − dV )
 √
A solid cylinder of radius R Charge q on a surface of conducting • 1 eV (electron volt) = 1.602 × 10−19 J A collection of point charges

1 Q  a2 + x2 + a 
sphere of radius R • ⃗E is in the direction from high V to Xq V = ln  √
U 1 X qi 4πϵϵ0 2a  a2 + x2 − a 

Inside sphere (r < R) ⇒ E = 0 i
low V V = = ke =
Outside sphere (r > R), then q0 ri 4πϵ0 ri
Electric potential energy i i Due to a finite horizontal line of
1 q Electric potential energy Due to a dipole charge
E=
4πϵ0 r 2
U = qV
Infinite conducting cylinder with ra-
dius R and charge per unit length λ
Change in electric potential energy
The point is at distance r from the center Inside cylinder (r < R) ⇒ E = 0
of the cylinder. ∆U = q∆V = q(Vf − Vi )
Outside cylinder (r > R), then
λr λ U : joule ; V : volt ; q: coulomb.
(r < R) (r > R) 1 λ
2πϵϵ0 R 2 2πϵϵ0 r E = Work done by the field
2πϵ0 r
Sphere with a hole W = Kf − Ki = −q(Vf − Vi )
A solid insulating sphere inside a For point P on the y-axis
A hollow sphere made from a non- spherical shell
conducting material is shown below in Work done by an applied force (ex-  
cross-section. The inner radius is R1 and ternal agent)  q −q 
VP = ke  p
 +p  = 0
outer radius R2 . The material is charged Wapp = (Kf − Ki ) + q(Vf − Vi ) a2 + y 2 a2 + y 2
uniformly ρ.
For point R on the positive x-axis
NOTE: When the particle is stationary √
before and after the move (Ki = Kf ): 2k qa
 
VR = − 2 e 2 Q  ℓ + a2 + ℓ 2 
x −a V = ke ln 
ℓ a

Wapp = q(Vf − Vi )
For x ≫ a
Work done by an external force to NOTE: If ℓ ≪ a, V = ke Q/a.
2k qa
move q from infinity to an electric VR ≈ − e2 Due to a hollow sphere
For r < R1 < R2 , E = 0 3 x
For R1 < r < R2 field created by q1 and q2
! Due to two oppositely charged par-
R31  q3 q1 q2 allel plates
 
4π  The conducting shell creates zero field W =U = +
E = ke ρ r − 2  4πϵ r r
3 r inside itself, so it does not affect regions 0 13 23
1 and 2.
For R1 < R2 < r Electric potential
Q For points inside the sphere
3
ρ(R2 − R1 )3 E 1 = k e r (for r < a) Finding V from E ⃗
E= a3 q
3r 2 ϵϵ0 Zf VM =
Q ⃗ · d⃗s 4πϵϵ0 R
E2 = ke 2 (for a < r < b) V f − Vi = − E
Conducting objects in an electric r i For points outside the sphere at a dis-
field Region 3 is a conductor in equilibrium, Obtaining the value of electric field tance r from its center
ADDITIONAL NOTES: therefore Set y = 0 and U = 0 at the bottom plate. q
from electric potential VN =
E3 = 0 (for b < r < c) U (y) q0 E(y) 4πϵϵ0 r
• The cases of a hollow cylinder or ∂V ∂V ∂V V (y) = = = Ey
sphere are similar to that of a solid In region 4, where r > c, construct a Ex = − Ey = − Ez = − q0 q0 For points near the surface (as the poten-
cylinder or sphere, except electric spherical gaussian surface; this surface ∂x ∂y ∂z tial on the surface is indeterminate)
fields inside such objects are zero. Due to a finite vertical line of
surrounds a total charge Qencl = Q + When E ⃗ is uniform q
• For two concentric spheres (or two (−2Q) = Q. Therefore, model the charge charge V =
∆V 4πϵϵ0 R
coaxial cylinders) with surface distribution as a sphere with charge −Q. E=−
charge densities σ1 and σ2 : ∆s Due to an infinite line charge or
– If the point is outside of the inner Q
E4 = −ke 2 (for r > c) Electric potential due to various spe- charged conducting cylinder
sphere (cylinder) and inside of r
the outer sphere (cylinder), only NOTE: If the sphere is conducting, cial cases
the inner surface creates E, Qencl = 0 for a gaussian surface of radius Due to point charges
– If the point is inside both spheres r < a ⇒ E1 = 0. A single point charge q
(cylinders), E is zero.
U q 1 q
– If the point is outside both ELECTRIC POTENTIAL V =
q0
= ke =
r 4πϵ0 r
spheres (cylinders), both sur- Some concepts
faces create E. Use superposition
to determine total E. • Potential difference ∆VAB = VA − VB
 Definition of capacitance Capacitors with dielectrics where ∆ρ = ρ−ρ0 is the resistivity change hence kinetic energy: K = 1 mv 2 =
λ r
V = ln 0 2
2πϵϵ0 r Q Capacitors half-filled with dielec- in the temperature interval ∆T = T − T0 . q2 B2 R2
C= 2m
Vab
tric material MAGNETIC FIELDS
NOTE: If we choose r0 to be the cylinder Spherical capacitors
of radius R, so that V = 0 when r = R, where Vab is the potential difference be- - Vector expression for the magnetic force
then at any point for which r > R, tween the two conductors. 2πϵ 0 (ϵ + 1)R 2 R1 1 on a charged particle moving in a mag-
C= = C0 (ϵ + 1) netic field:
Parallel-plate capacitor R 2 − R 1 2 −−→ →
−
λ R
V = ln FB = q→ −
v ×B
2πϵϵ0 r Two parallel, metallic plates of equal Cylindrical capacitors
area A are separated by a distance d. - Magnitude of that magnetic force:
Inside the cylinder, ⃗E = 0, and V has the πϵ0 (ϵ + 1)l 1 FB = |q| vB sin θ
ϵϵ A C=  R  = C0 (ϵ + 1)
same value as on the cylinder’s surface. C= 0 ln R2 2 Charge moving in circular path
d 1
Due to a uniformly charged ring
where C0 is the original capacitance of
Spherical capacitor the capacitor and l is the length of the Magnetic Force Acting on Current-
capacitor. Carrying Conductor
Dielectric partially filling the gap - Average magnetic force exerted on a
in a capacitor →
−
charge moving in a wire: q −
v−→
d ×B

- Magnetic force on wire segment of

ke Q A spherical conducting shell of radius b −−→ → − → −
V =√ - Magnetic force on a charge moving in length L: FB = I L × B
and charge −Q concentric with a smaller
a2 + x2 mv 2
conducting sphere of radius a and charge circular motion: FB = qvB = (accel- - Total force acting over the length of the
Due to a uniformly charged disk Q. r −−→ Rb − → −
eration > centripetal acceleration) wire: FB = I a d→ s × B , a & b represent
ab mv
C= - Radius of the circular path: r = endpoint of the wire.
ke (b − a) qB
ϵϵ0 A Torque on a Current Loop
C′ = - Angular speed of the particle:
When the b → ∞, the inner sphere be- ϵd + (1 − ϵ)b v qB
comes a point charge ⇒ The system be- ω= =
r m
comes a conducting shell with a point CURRENT & RESISTANCE - Period of the motion:
charge inside. 2πr 2π 2πm
Current density T = = =
p  C = 4πϵϵ0 a Defined as the current per unit area. v ω qB
V = 2πke σ R2 + x2 − x Cylindrical capacitor I Some applications - Torque on current loop in a magnetic
J≡ = nqvd = σ E →
− → − →−
A - Total force (Lorentz force) acting on the field: →
−
τ = I A × B , where A ⊥ the loop.
Potential difference between two, charge: →
−
The constant of proportionality σ is the −→ →
− →
− - The product I A is "magnetic dipole
in equal magnitude, oppositely conductivity of the conductor. F = q E + q→−
v ×B moment → −
µ ": →
−
µ =IA
→
−
charged concentric spherical sur- E →−
- Velocity Selector: v =
faces NOTE: The inverse of conductivity B - If a coil contains N loops: →
−
µ = NI A
σ is called resistivity ρ = 1/σ - Torque on a magnetic moment: → −τ =
→− →
−
Q(R2 − R1 ) Resistance of uniform material along µ ×B
U = V1 − V2 = length ℓ - Potential energy of a system of a mag-
4πϵϵ0 R1 R2 L
C= netic moment in a magnetic field: UB =
Potential difference between two, 2ke ln (rb /ra ) ℓ →− →
−
R=ρ µ ·B
in equal magnitude, oppositely Energy storage in a charged capaci- A
Hall Effect
charged coaxial cylindrical sur- tor Variation of resistance with tempera- - Mass spectrometer: m = rB0 = rB0 B
q v E
faces
Energy stored inside a capacitor ture
(used for all types of capacitors) ρ = ρ0 [1 + α (T − T0 )]
λ R
U = V1 − V2 = ln 2 R = R0 [1 + α (T − T0 )]
2πϵϵ0 R1 Q∆V C(∆V )2 Q2
W= = = where α is the temperature coefficient of
2 2 2C
CAPACITANCE resistivity.
Electric field inside a capacitor
Dielectric constant 1 ∆ρ
σ q 1 α= - The Hall voltage
κ≡ϵ ρ0 ∆T
E= = qBR IB R IB
ϵϵ0 S ϵϵ0 - Cyclotron: we known that v = m , ∆VH = = H
nqt t
where |d⃗s × r̂| ds µo N I - If the magnetic field is uniform every- Lenz’s Law
HINT: = 2 2 B =
r2 a +s 2πr where
• t = thickness Magnetic force between 2 parallel Magnetic field of a solenoid d
• n = charge-carrier density E = −N (BA cos θ)
conductors dt
• q = charge
• RH = nq1 = Hall coefficient A: area of the loop
Motional emf ⃗ ext
SOURCES OF THE - When the copper bar moves inside a
dB ⃗ int
↑↓ B
dt
MAGNETIC FIELD uniform magnetic field, emf is created
between 2 end of the bar. ⃗ ext : magnetic field created by the mag-
-B
Biot-Savart Law - The bar then acts like a Voltage source. net.
µo Id⃗s × r̂ ⃗ int : magnetic field created by the cur-
-B
⃗ =
dB
4π r2 rent in the loop.
µo = 4π × 10−7 (T · m/A) : permeability of µo I1 I2 N Induced emf and electric field
F1 (or F21 ) = I1 l B2 = B = µo I = µo nI
free space (NOT permittivity) 2πa l
l: the entire length of the solenoid (m)
I
⃗
d⃗s : a small length of wire with d⃗s ↑↑ I F B µ o I 1 I 2 ⃗ · d⃗s = − dΦB
= N : number of loops E
⃗r : a vector from d⃗s to point P l 2πa dt
r̂ : a unit vector with the same direction Ampere’s law n = N /l: number of loops over 1 unit
length (loops/m) Solenoid
as ⃗r - If the current I is constant everywhere
µ I
Z
d⃗s × r̂ Magnetic flux Let say we have a solenoid with
B⃗ = o in the wire radius R, I = Im cos(ωt)
r2
I
4π
I I
⃗ · d⃗s = B ds = µo I
B ΦB = B ⃗
⃗ · dA
Magnetic field around a straight |E| = Blv
conductor - If the current I is not constant every- ΦB : magnetic flux (W b)
where in the (the direction of the current is
I wire
determined by using Lenz’s Law at III)
Z
⃗ · d⃗s = µo J dA
B - If the magnetic field is uniform
⃗ = BAcos(∠B,
⃗ ·A ⃗
⃗ A) - If B ⃗ B)
⃗ ̸⊥ v⃗ then E = Blv cos (∠A, ⃗ with
where J is the current density. ΦB = B
⃗
Long current carrying wire Magnetic field through a rectangu- A is the area of the loop.
- Energy transferred to the bar by thes
µo I
lar loop ⃗app is
B= (sin θ1 − sin θ2 ) constant force F
4πa B2 l 2 v 2 E2 µo nIωR2
- Note that θ1 > 0 and θ2 < 0 P = Fapp v = (IlB)v = = - if r > R then E = 2r sin(ωt)
- HINT: d⃗s × r̂ = (dxcos θ)k̂ R R
µo nIm ω
Magnetic force acting on a sliding - if r < R then E = r sin(ωt)
Magnetic field around a curved 2
bar
wire ⃗app on the - Alternate formula:
- If there is no constant force F
µo I bar. Instead the bar is only given an ini- −µo nR2 dI
B = (for r > R) tial velocity v⃗i . The equation of v⃗ as a E = × (r > R)
H 2πr H µo Ib a function of time is 2r dt
⃗
- HINT: B · d⃗s = B ds = B × 2πr ΦB = ln(1 + ) v = vi e−t/τ −µ o nr dI
2π c E = × (r < R)
µo I mR 2 dt
B = r (for r < R) with τ = 2 2 ′
HINT for r > R: − (ΦB )t = −(BA)′t =
2πR 2
µo I
R a+c B l
−πR2 (µo nI)′t
R
- HINT: dA = c b dr
H H
B= θ - HINT: B ⃗ · d⃗s = B ds = B(2πr) = Motional emf induced in a rotating
4πa 2 Generators and Motors
R
HINT: ds = s = aθ µo I ′ = µo I r 2
R
FARADAY’S LAW bar
- emf generated by 1 loop of wire spin-
Magnetic field of a current loop Toroid Faraday’s law of induction ning inside the magnetic field:
- emf: electromotive force E = BAω sin(ωt)
- When the flux through N number of - emf generated by N loop of wire
close loop changes ⇒ emf is created E = N BAω sin(ωt)
dΦB
E =−N
dt

N : number of loops
ΦB : magnetic flux (W b)
µo Ia2 t: time (s) 1
B = Bx = E: emf (V ) E = Bωl 2
2(a + x2 )3/2
2
2