University Physics Equation Sheet

PH2200 Electric Charges and Forces

K q1 q2 F1 on 2 = F2 on 1 = r2 q = ( N p − Ne ) e G G E = Fon q / q G G Fon B = qB E G 1 q E= rˆ point charge 4πε 0 r 2

Formula Sheet Potential and Field

Parallel-plate capacitor: G ⎛η ⎞ E = ⎜ , from positive to negative ⎟ ⎝ ε0 ⎠ ( Ering ) z = 4πε1 2 zQ2 3/2 0 (z + R )

( Edisk ) z =

η 2ε 0

G G a = (q / m) E

⎡ ⎤ z ⎢1 − 2 ⎥ z + R2 ⎦ ⎣

i

Electric dipole: G p -q + +q s G p = ( qs, from negative to positive ) G G 1 2p Field on axis E = 4πε 0 r 3 G G 1 p Field in bisecting plane E = − 4πε 0 r 3

τ = pE sin θ

Q L Q surface charge density :η = A Q volume charge density : ρ = V Uniform infinite line of charge:

G G Φ e = E ⋅ A = EAcosθ constant field G G Φ e = ∫ E ⋅ dA surface

G G Q Φ e = v∫ E ⋅ dA = in

ε0

G E at surface of a charged conductor: ⎧η ⎪ = ⎨ε0 ⎪perpendicular to surface ⎩

linear charge density : λ =

G ⎛ 1 2λ ⎞ E =⎜ , perpendicular to line ⎟ ⎝ 4πε 0 r ⎠ Uniform infinite plane of charge: G ⎛ η ⎞ E =⎜ , perpendicular to plane ⎟ ⎝ 2ε 0 ⎠ Uniformly charged sphere: G Q E= rˆ for r ≥ R 4πε 0 r 2

si

= the negative of the "area" dV ds = ∑ ( ΔV ) i = 0

Es = − ΔVloop

i

C=

Gauss’s Law

Esurface

sf

ΔV = V ( sf ) − V ( si ) = − ∫ Es ds

W ΔVbat = chem = E (ideal battery) q Q C= C = κ C0 ΔVC

The Electric Field

G G Enet = ∑ Ei

Knight

ε0 A

(parallel-plate capacitor) d Ceq = C1 + C2 + C3 + ... (parallel) −1

⎛ 1 ⎞ 1 1 Ceq = ⎜ + + + ... ⎟ (series) ⎝ C1 C2 C3 ⎠ 2 ε Q 1 2 = C ( ΔVC ) UC = uE = 0 E 2 2C 2 2 Current and Conductivity

Electron current: i = rate of electron flow N e = iΔt i = nAvd

The Electric Potential U elect = U 0 + qEs (parallel-plate capacitor) 1 q1q2 1 qi q j U elect = ∑ 4πε 0 r i < j 4πε 0 rij G G = − p ⋅ E = − pE cos θ

U q1 + q2 = U dipole

U q +sources = qV V = Es

V=

U q +sources q

( inside a parallel-plate capacitor )

ΔVC (parallel plate capacitor) d 1 q point charge V= 4πε 0 r E=

V =∑ i

1 qi 4πε 0 ri

eτ E m Conventional current: I = rate of charge flow = ei vd =

Q = I Δt Current density: J =I/A J = nevd = σ E ne 2τ 1 = ρ m

σ=

∑I

in

= ∑ I out

Ewire = I=

ΔVwire L

ΔVwire ρL where R = R A

Need additional help? Then visit the Physics Learning Center located in 228 Fisher. Walk-in hours are Monday 7:00 – 9:00 p.m., and Tuesday through Thursday 3:00 - 9:00 p.m.

Fundamentals of Circuits

ΔV R junction law :

I=

∑I

in

Useful Geometry

dt

Ecoil = ω ABN sin ωt

= ∑ I out

ΔVloop = ∑ ( ΔV )i = 0

loop law :

d Φ per turn

E=N

V2 =

N2 V1 N1

i

Pbat = I E

Electromagnetic Fields and Waves

PR = I ΔVR = I 2 R =

( ΔVR )

2

R Req = R1 + R2 + R3 + ... + RN (series) −1

⎛ 1 1 1 1 ⎞ Req = ⎜ + + + ... + ⎟ (parallel) RN ⎠ ⎝ R1 R2 R3 Q = Q0 e − t / τ I = I 0 e− t / τ τ = RC The Magnetic Field

G μ B= 0 4π G μ B= 0 4π

G ⎞ qv × rˆ ⎛ μ0 q v sin θ =⎜ , RHR ⎟ 2 2 r ⎝ 4π r ⎠ G ⎞ I Δs × rˆ ⎛ μ0 I ( Δs ) sin θ =⎜ , RHR ⎟ 2 2 4π r r ⎝ ⎠

Blong straight wire

μ I = 0 2π d

Bcoil center =

μ 0 NI

2 R μ = ( AI , from south pole to north pole ) G G μ 2μ Bdipole = 0 3 (on axis of dipole) 4π z G G v∫ B ⋅ ds = μo I through G

G

G

μo NI

Bsolenoid = L G G G Fon q = qv × B = ( q vB sin θ , RHR ) qB mv rcyc = qB 2π m G G = IL × B = ( ILB sin θ , RHR )

Qin

v∫ E ⋅ dA = ε G G v∫ B ⋅ dA = 0 G

G

v∫ E ⋅ ds = −

0

d Φm dt

G G dΦ B v∫ ⋅ ds = μ0 I through + ε 0 μ0 dt e G G G G F =q E+v×B

(

)

d Φe dt = c = 1 / ε 0 μ0

I disp = ε 0 vem

c=λf

μ LI I Fparallel wires = 0 1 2 2π d G G G τ = μ × B = ( μ B sin θ , RHR )

Cylinder Lateral surface area = 2π rL Volume = π r 2 L PH2100 in Brief

G G G Fnet = ∑ Fi = ma i

G G FA on B = − FB on A Fspring = − k Δs

(

xf = xi + vix Δt + 12 ax ( Δt )

)

cε P 1 I = = Savg = E02 = 0 E02 A 2c μ 0 2 F I prad = = (perfect absorber) A c I = I 0 cos 2 θ I transmitted =

1 I 0 (incident light unpolarized) 2

f cyc = G Fwire

Sphere Surface area = 4π r 2 Volume = 43 π r 3

Constant Acceleration :

E = cB G 1 G G S= E×B

μ0

Circle Area = π r 2 Circumference = 2π r

Physical Constants

K = 8.99 × 109 N ⋅ m 2 /C 2

ε 0 = 8.85 × 10−12 C 2 /N ⋅ m 2 e = 1.60 × 10−19 C

Electromagnetic Induction

me = 9.11 × 10−31 kg

E = vlB G G G Φ m = A ⋅ B = AB cos θ (uniform B-field) G G Φ m = ∫ B ⋅ dA

mp = 1.67 × 10−27 kg

μ0 = 1.26 × 10−6 T ⋅ m/A c = 3.00 × 108 m/s

2

yf = yi + viy Δt + 12 a y ( Δt )

2

vfx = vix +ax Δt vfy = viy + a y Δt vf2x = vi2x + 2ax ( xf − xi ) vf2y = vi2y + 2a y ( yf − yi ) Uniform Circular Motion : 2π r 2π rad ω= T T θ f = θi + ω Δt v=

ar =

v2 = ω 2r r

Energy Conservation K = 12 mv 2 Emech = K + U Kf + U f = Ki + U i

G G P = F ⋅ v = Fv cos θ

area of loop

Need additional help? Then visit the Physics Learning Center located in 228 Fisher. Walk-in hours are Monday 7:00 – 9:00 p.m., and Tuesday through Thursday 3:00 - 9:00 p.m.