Physics 1302W
Final Examination
May 10, 2025
This is a closed-book exam. No outside materials are allowed except for a calculator, which
may not be used as a memory-storage or communication device.
Be sure to write your full name, x500 (e.g. stud0543), and fill in the bubbles and/or text of
your student ID (e.g., 5454545) on all pages of the answer sheet template. Also be sure to
fill in the bubbles on the last page of the answer sheets with your answers to the multiple
choice problems.
Answers to each of the long answer problems should be completed on the identifying page of
the answer sheets, using the backside if more space is required. For these problems, graders
cannot give you any partial credit unless they can understand your methodology.
Some useful formulae:
Mechanics (from 1301):
P~
2
ac = vr (Centripetal acceleration)
F = m~a (Newton’s second law)
Ei + W = Ef (Work-energy theorem) E = K + U (Mechanical energy)
K = 21 mv 2 (Kinetic energy)
Electric Fields and Forces:
~ = k 2q r̂ (Electric field of a point charge)
F~12 = k rq21 q2 r̂12 (Coulomb force)
E
r
12
~ (Electric force on a charge q in external field)
F~ = q E
~ (Dipole torque)
~ (Dipole energy)
p~ = qRd~ (Dipole Moment) ~τ = p~ × E
U = −~p · E
~ = kr̂2 dq (Electric field of a continuous charge distribution)
E
H
Rr
~ · dA
~ = Qenc (Gauss’s Law)
~ · dA
~ (Electric flux)
ΦE = E
ΦE = E
S
S
ǫ0
Electric Potential:
Rf
~ · d~l (Electric Potential Difference)
∆V = Vf − Vi = − i E
R
(Potential of a point charge) V = k rdq (Potential of finite continuous distribution)
V = kq
r
~ = −∇V
~ (Relationship between field and potential)
Er = − dV
or E
dr
U = qV (Potential energy of a charge q in a potential V )
U = 21 Σqi Vi (Energy of series of point charges)
U = 21 QV (Self-energy of charged conductor)
Capacitance:
2
C = Q/V
C = ǫ0 Ad (Parallel plate capacitance) U = 12 CV 2 = 21 QC (Energy of a capacitor)
ue = 21 ǫ0 E 2 (Energy density of electric field in vacuum)
Ceq = C1 + C2 + ... (Capacitors in parallel)
1/(Ceq ) = 1/C1 + 1/C2 + ... (Capacitors in
series)
Physics 1302
Final Exam
May 10, 2025
1
E = Eκ0 (Field in dielectric placed in an external field E0 ) ǫ = κǫ0 (Permittivity of dielectric)
C = κC0 (Capacitance of a capacitor completely filled with dielectric)
Resistance
R and circuits:
~ (Current definition where J = nevd , n = number density of charge carriers)
=
J~ ·dA
i = ∆Q
∆t
R = ρ(L/A) (Resistance-resistivity relation)
ρ − ρ0 = ρ0 α(T − T0 ) (Resistivity-temperature relation)
ρ = nem2 τ (resistivity in metal, τ = mean time between collisions)
2
V = iR (Ohm’s Law) P = iV = i2 R = VR (Resistor power)
P = iE (Battery power)
Req = R1 + R2 + ... (Resistors in series)
1/Req = 1/R1 + 1/R2 + ... (Resistors in parallel)
Kirchhoff’s rules: (1) Sum of potentials/voltages around closed loop in circuit = 0.
(2) Net current flowing into node = 0.
RC circuits: Charging a capacitor through single resistor Q = Qf (1 − e−t/τ ) and i = i0 e−t/τ
where τ = RC, i0 = E/R
Discharging a capacitor through single resistor Q = Q0 e−t/τ and i = i0 e−t/τ where i0 = V0 /R
Magnetism:
~
~ ×B
~
~ (B-force on moving charge/current)
F~ = q~v × B
F~ = i L
dF~ = i d~l × B
~
~
~
µ
~ = N iA
~τ = ~µ × B
U = −~µ · B
(Magnetic dipole)
~
µ0 idl×r̂
~
dB = 4π r2 (Biot-Savart Law for B-field of current piece)
H
~ · d~l = µ0 ienc (Ampere’s Law)
B
C
Electromagnetic
Induction:
R
~
~
ΦB = S B · dA (Magnetic flux through a surface S)
H
~ · d~s and N = number of loops (Faraday’s Law)
E = −N dΦdtB where E = C E
di
E = −L dt
where L = N Φi B (Inductor)
B2
UB = 12 Li2 (Energy of inductor)
uB = 2µ
(B-field energy density)
0
E
−t/τL
) where τL = L/R (Rise of DC current in inductor)
i = R (1 − e
−t/τL
i = i0 e
(Decay of DC current in inductor)
di1
E2 = −M dt (Mutual inductance)
AC Circuits:
Irms = √12 Imax (for sinusoidal current, same for voltage)
VL = IXL , VL = IXC , XL = ωd L, XC = ωd1C (inductive and capacitive reactance)
For RLC series circuits: If E = Emax cos(ωd t) then i = Emax
cos(ωd t − φ)
Z
2
2
Pave = Irms
Rp= Irms
Z cos φ
C
where Z = R2 + (XL − Xc )2 , tan φ = XL −X
R
NS
VS
= IIPS = N
(Transformer)
VP
P
EM
Waves and Maxwell’s Equations:
H
R
~l = µ0 (ienc + id ) = µ0 ienc + µ0 ǫ0 d
~
~ · dA
~ (Ampere’s law with Maxwell fix)
B
·
d
E
dt S
C
Physics 1302
Final Exam
May 10, 2025
2
Erms Brms
0 B0
~
I = Power
= E2µ
(Average intensity of EM wave, where
µ0
0
Area = |S|ave = uave c =
E = Bc)
λf = ωk = c (where k = 2π/λ and ω = 2πf = 2π/T )
E = Emax sin(kx − ωt) (Electric field magnitude of sinusoidal plane wave traveling in +x
direction)
R
~ B
~
~ = E×
~ · dA)
~
S
(Poynting vector, where energy flux [i.e., power] P = s S
µ0
p = Uc (Momentum of EM wave)
Pr = Ic (Radiation pressure, w/ 100% absorption, of EM wave)
I = I0 cos2 θ (Law of Malus for polarized light)
cvac
, cvac = 3.0 × 108 m/s)
n1 sin θ1 = n2 sin θ2 (Snell’s law of refraction where n = cmedia
constants:
g = 9.8 m/s2
e = 1.6 × 10−19 C
1
k = 4πǫ0 = 9.0 × 109 N·m2 /C2
ǫ0 = 8.85 × 10−12 C2 /(N·m2 )
µ0 = 4π × 10−7 T·m/A (vacuum permeability)
c = √µ10 ǫ0 = 3.0 × 108 m/s (Speed of light in vacuum)
√
2
If ax2 + bx + c = 0, then x = −b± 2ab −4ac .
~ ~
~ × B|
~ = AB sin θ
A
θ = Ax Bx + Ay By + Az Bz
|A
R · nB = AB1cosn+1
+ C forRn 6= −1 and ln |x| + C for n = −1
R x dx = n+1 x
cos θ dθ = sin θ + C and sin θ dθ = − cos θ + C
Physics 1302
Final Exam
May 10, 2025
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