Useful Equations for PHYS140 Exam 3
2
q1 q2
1 q1 q2
1
F~ = k 2 =
r̂ where k =
= 9 × 109 NCm2
2
r
4π0 r
4π0
~
~ =F
E
q
F~net = ΣF~i
1 2~
p
on axis
=
4π0 r3
~ dipole
E
~ dipole =
E
1 p~
perpendicular plane
4π0 r3
1 2λ
4π0 r
~ disk | = η [1 − √ z
|E
]
20
z 2 + R2
~ sphere | = 1 Q
|E
4π0 r2
η
z>0
20
~ plane =
E
η
− 20 z < 0
~ line | =
|E
~ capacitor | =
|E
η
0
~ ·A
~ (electric flux through a flat area)
φ=E
ˆ
~ · dA
~ (electric flux through a surface)
φ= E
˛
~ · dA
~ = Qin (electric flux through a closed surface)
E
0
U = U0 + qEs
U =k
ie = ne Avd
ne eτ A
E
m
I
= ne evd
J=
A
ΣIin = ΣIout
ie =
~ = k q r̂ = 1 q r̂
E
r2
4π0 r2
φ=
Q
0 A
E=
1 q1 q2
q1 q2
=
r
4π0 r
~
Udipole = −pEcosφ = −~
p·E
ne e2 τ
m
J = σE
1
m
ρ= =
σ
ne e2 τ
σ=
∆V
R
Pbat = I
I=
W = F ∆s = qEd
(∆VR )2
R
= R1 + R2 + R3 + ... series resistors
PR = I∆VR = I 2 R =
Req
Req = (
Q = Q0 e−t/τ
~ pointcharge = µ0 q~v × r̂
B
4π r2
~ currentsegment = µ0 I∆~s × r̂
B
4π
r2
µ0 N I
Bcoilcenter =
2 R
µ
~ = AI
˛
~ · d~s = µ0 I
E
U = qV
V = Es inside parallel-plate capacitor
1 qi
qi
= Σk
4π0 r
r
ˆ
~ · d~s
∆V = Vf − Vi = − E
V =Σ
dV
ds
= Σ(∆V )i = 0
Es = −
∆Vloop
Q = C∆Vc
Ceq = C1 + C2 + C3 + ... parallel capacitors
Ceq = (
C=
1
1
1
+
+
+ ...)−1 series capacitors
C1
C2
C3
Q
0 A
=
∆Vc
d
1
1
1
+
+
+ ...)−1 parallel resistors
R1
R2
R3
µ0 N I
= µ0 nI
I
~
= q~v × B
Bsolenoid =
F~onq
~
F~wire = lI~ × B
fcyc =
qB
2πm
µ0 I2
µ0 lI1 I2
F~wires = I1 lB2 = I1 l
=
2πd
2πd
dφm
=|
|
dt
ˆ
1
UL = L IdI = LI 2
2
~0 = E
~ +V
~ ×B
~
E
~0 = B
~− 1V
~ ×E
~
B
c2
~ =E
~0 − V
~ ×B
~0
E
~ =B
~0 + 1 V
~ ×E
~0
B
c2
~0 = B
~+ 1V
~ ×E
~
B
c2
˛
~ · d~s = µ0 I + 0 µ0 dφe
B
dt
D(x, t) = A sin (kx − ωt + φ0 )
n= vc
λmat =
P
I= Area
β = (10 dB) log10
~ + ~v × B)
~
F~ = q(E
vem = √
1
= 3.00 × 108 m/s = c
0 µ0
~ + ~v × B)
~
F~ = q(E
I=
1
P
c0 2
= Savg =
E02 =
E
A
2cµ0
2 0
P
Intensity
F
=
=
A
cA
c
prad =
λvac
n
I
I0
D(x,q
t) = A (x) cos (ωt)where A (x) = 2a sin (kx)
v = Tµs
f+ =
f0
1−vs /v
for an approaching source
f− =
f0
1+vs /v
for a receding force
- for an observer approaching a source:
Itransmitted = I0 cos2 θ
f+ = (1 + v0 /v) f0
F
P
Intensity
=
=
A
cA
c
prad =
- for an observer approaching a source:
f− = (1 − v0 /v) f0

0

x = x + vt
0
y=y

0

z=z
∆t = √∆τ
1−β 2
0
L =
p
- for standing waves on a string:
(
λm = 2L
m
v
v
fm = λvm = 2L/m
= m 2L
≥ ∆τ
1 − β2l ≤ l
2
- for open-open or closed-closed tube:
(
λm = 2L
m,
m = 1, 2, . . .
v
fm = m 2L
= mf1
2
s2 = c2 (∆t) − (∆x)
γ=√
m = 1, 2, . . .
1
1−v 2 /c2
=√
1
1−β 2
 0
x = γ (x − vt)



y 0 = y
0

z =z


0
t = γ t − vx/c2
- for open-closed tube:
0
u =
u−v
1−uv/c2
(
λm = 4L
m,
v
fm = m 4L
= mf1
m = 1, 2, . . .
- for maximum constructive interference:
p = γp mu
E = γp mc2 = E0 + K
2
E 2 = (pc) = E02
∆φ = 2π
∆x
+ ∆φ0 = m · 2π
λ
- for maximum destructive interference:
m
L
µ=
T= f1
∆x
∆φ = 2π
+ ∆φ0 =
λ
v= Tλ = λf
k=
2π
λ
ω = vk
mod
fbeat = 2fmod = 2 ω2π
= f1 − f2
1
m+
· 2π
2
θm = m λd angles of bright fringes
ym =
mλL
d
positions of bright fringes
y0 m = m + 12 λL
d positions of dark fringes
πd
Idouble = 4I1 cos2 λL
y
d sinθm = mλdark fringes
ym = L tanθm bright fringes
θp = p λa , p = 1, 2, . . . angles of dark fringes
yp =
pλL
a ,
w=
2λL
a
p = 1, 2, . . . position of dark fringes
w = 2y1 = 2L tanθ1 ≈
θ1 = sin−1 λa
2.44λL
D
Constants
c = 3.00 × 108 m/s
µ0 = 1.26 × 10−6 T m/A
ε0 = 8.85 × 10−12 C 2 /N m2
Charge Densities
Q
L
Q
η=
A
λ=
ρ=
Q
V olume