2.5 Electric charges and fields
ε 0 = {\displaystyle \varepsilon _{0}=} 8.85×10−12 F /m = vacuum permittivity.
e = 1.602×10−19 C : negative (positive) charge for electrons (protons)
k e = 1 4 π ε 0 = {\displaystyle k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=} = 8.99×109 m/F
F → = Q E → {\displaystyle {\vec {F}}=Q{\vec {E}}} where E → = 1 4 π ε 0 ∑ i = 1 N q i r P i 2 r ^ P i {\displaystyle {\vec {E}}={\tfrac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Pi}^{2}}}{\hat {r}}_{Pi}}
E → = ∫ d q r 2 r ^ {\displaystyle {\vec {E}}=\int {{\tfrac {dq}{{r}^{2}}}{\hat {r}}}} where d q = λ d ℓ = σ d a = ρ d V {\displaystyle dq=\lambda d\ell =\sigma da=\rho dV}
E = σ 2 ε 0 {\displaystyle E={\tfrac {\sigma }{2\varepsilon _{0}}}} = field above an infinite plane of charge.
Find E
E → ( r → ) = 1 4 π ε 0 ∑ i = 1 N R ^ i Q i | R → i | 2 = 1 4 π ε 0 ∑ i = 1 N R → i Q i | R → i | 3 {\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}} is the electric field at the field point, r → {\displaystyle {\vec {r}}} , due to point charges at the source points,r → i {\displaystyle {\vec {r}}_{i}} , and R → i = r → − r → i , {\displaystyle {\vec {\mathcal {R}}}_{i}={\vec {r}}-{\vec {r}}_{i},} points from source points to the field point.
2.6 Gauss's law :
Φ = E → ⋅ A → {\displaystyle \Phi ={\vec {E}}\cdot {\vec {A}}}
→ ∫ E → ⋅ d A → = ∫ E → ⋅ n ^ d A {\displaystyle \to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA} = electric flux
q e n c l o s e d = ε 0 ∮ E → ⋅ d A → {\displaystyle q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}}
d Vol = d x d y d z = r 2 d r d A {\displaystyle d\,{\text{Vol}}=dxdydz=r^{2}drdA} where d A = r 2 d ϕ d θ {\displaystyle dA=r^{2}d\phi d\theta }
A sphere = r 2 ∫ 0 π sin θ d θ ∫ 0 2 π d ϕ = 4 π r 2 {\displaystyle A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}}
Surface Integrals
Calculating ∫ f d A {\displaystyle \int fdA} and ∫ f d V {\displaystyle \int fdV} with angular symmetry
Cyndrical: d A = 2 π r d z ; d V = d A d r {\displaystyle dA=2\pi r\,dz;\,dV=dA\,dr} . Spherical: ∫ d A = 4 π r 2 , d V = 4 π r 2 d r {\displaystyle \int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr}
More Gauss Law
Calculating ∫ f d A {\displaystyle \int fdA} and ∫ f d V {\displaystyle \int fdV} with angular symmetry
Cyndrical: d A = 2 π r d z ; d V = d A d r {\displaystyle dA=2\pi r\,dz;\,dV=dA\,dr} . Spherical: ∫ d A = 4 π r 2 , d V = 4 π r 2 d r {\displaystyle \int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr}
2.7 Electric potential The alpha-particle is made up of two protons and two neutrons.
Δ V A B = V A − V B = − ∫ A B E → ⋅ d ℓ → {\displaystyle \Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}} = electric potential
E → = − ∂ V ∂ x i ^ − ∂ V ∂ y j ^ − ∂ V ∂ z k ^ = − ∇ → V {\displaystyle {\vec {E}}=-{\tfrac {\partial V}{\partial x}}{\hat {i}}-{\tfrac {\partial V}{\partial y}}{\hat {j}}-{\tfrac {\partial V}{\partial z}}{\hat {k}}=-{\vec {\nabla }}V}
q Δ V {\displaystyle q\Delta V} = change in potential energy (or simply U = q V {\displaystyle U=qV} )
P o w e r = Δ U Δ t = Δ q Δ t V = I V = e Δ N Δ t {\displaystyle Power={\tfrac {\Delta U}{\Delta t}}={\tfrac {\Delta q}{\Delta t}}V=IV=e{\tfrac {\Delta N}{\Delta t}}}
Electron (proton) mass = 9.11×10−31 kg (1.67× 10−27 kg). Elementary charge = e = 1.602×10−19 C.
K = 1 2 m v 2 {\displaystyle K={\tfrac {1}{2}}mv^{2}} =kinetic energy . 1 eV = 1.602×10−19 J
V ( r ) = k q r {\displaystyle V(r)=k{\tfrac {q}{r}}} near isolated point charge
Many charges: V P = k ∑ 1 N q i r i → k ∫ d q r {\displaystyle V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}} .
2.8 Capacitance
Q = C V {\displaystyle Q=CV} defines capacitance .
C = ε 0 A d {\displaystyle C=\varepsilon _{0}{\tfrac {A}{d}}} where A is area and d<<A1/2 is gap length of parallel plate capacitor
Series : 1 C S = ∑ 1 C i . {\displaystyle {\text{Series}}:\;{\tfrac {1}{C_{S}}}=\sum {\tfrac {1}{C_{i}}}.} Parallel: C P = ∑ C i . {\displaystyle {\text{ Parallel:}}\;C_{P}=\sum C_{i}.}
u = 1 2 Q V = 1 2 C V 2 = 1 2 C Q 2 {\displaystyle u={\tfrac {1}{2}}QV={\tfrac {1}{2}}CV^{2}={\tfrac {1}{2C}}Q^{2}} = stored energy
u E = 1 2 ε 0 E 2 {\displaystyle u_{E}={\tfrac {1}{2}}\varepsilon _{0}E^{2}} = energy density
2.9 Current and resistors
Electric current: 1 Amp (A) = 1 Coulomb (C) per second (s)
Current=I = d Q / d t = n q v d A {\displaystyle I=dQ/dt=nqv_{d}A} , where
( n , q , v d , A ) {\displaystyle (n,q,v_{d},A)} = (density, charge, speed, Area)
I = ∫ J → ⋅ d A → {\displaystyle I=\int {\vec {J}}\cdot d{\vec {A}}} where J → = n q v → d {\displaystyle {\vec {J}}=nq{\vec {v}}_{d}} =current density .
E → = ρ J → {\displaystyle {\vec {E}}=\rho {\vec {J}}} = electric field where ρ {\displaystyle \rho } = resistivity
ρ = ρ 0 [ 1 + α ( T − T 0 ) ] {\displaystyle \rho =\rho _{0}\left[1+\alpha (T-T_{0})\right]} , and R = R 0 [ 1 + α Δ T ] {\displaystyle R=R_{0}\left[1+\alpha \Delta T\right]} ,
where R = ρ L A {\displaystyle R=\rho {\tfrac {L}{A}}} is resistance
V = I R {\displaystyle V=IR} and Power=P = I V = I 2 R = V 2 / R {\displaystyle P=IV=I^{2}R=V^{2}/R}
2.10 Direct current circuits
2.13 Electromagnetic induction
2.14 Inductance
2.15 Alternating current circuits
AC voltage and current v = V 0 sin ( ω t − ϕ ) {\displaystyle v=V_{0}\sin(\omega t-\phi )} if i = I 0 sin ω t . {\displaystyle i=I_{0}\sin \omega t.} RMS values I r m s = I 0 2 {\displaystyle I_{rms}={\tfrac {I_{0}}{\sqrt {2}}}} and V r m s = V 0 2 {\displaystyle V_{rms}={\tfrac {V_{0}}{\sqrt {2}}}} Impedance V 0 = I 0 X {\displaystyle V_{0}=I_{0}X} Resistor V 0 = I 0 X R , ϕ = 0 , {\displaystyle V_{0}=I_{0}X_{R},\;\phi =0,} where X R = R {\displaystyle X_{R}=R} Capacitor V 0 = I 0 X C , ϕ = − π 2 , {\displaystyle V_{0}=I_{0}X_{C},\;\phi =-{\tfrac {\pi }{2}},} where X C = 1 ω C {\displaystyle X_{C}={\tfrac {1}{\omega C}}} Inductor V 0 = I 0 X L , ϕ = + π 2 , {\displaystyle V_{0}=I_{0}X_{L},\;\phi =+{\tfrac {\pi }{2}},} where X L = ω L {\displaystyle X_{L}=\omega L} RLC series circuit V 0 = I 0 Z {\displaystyle V_{0}=I_{0}Z} where Z = R 2 + ( X L − X C ) 2 {\displaystyle Z={\sqrt {R^{2}+\left(X_{L}-X_{C}\right)^{2}}}} and ϕ = tan − 1 X L − X C R {\displaystyle \phi =\tan ^{-1}{\frac {X_{L}-X_{C}}{R}}} Resonant angular frequency ω 0 = 1 L C {\displaystyle \omega _{0}={\sqrt {\tfrac {1}{LC}}}} Quality factor Q = ω 0 Δ ω = ω 0 L R {\displaystyle Q={\tfrac {\omega _{0}}{\Delta \omega }}={\tfrac {\omega _{0}L}{R}}} Average power P a v e = 1 2 I 0 V 0 cos ϕ = I r m s V r m s cos ϕ {\displaystyle P_{ave}={\frac {1}{2}}I_{0}V_{0}\cos \phi =I_{rms}V_{rms}\cos \phi } Transformer voltages and currents V S V P = N S N P = I P I S {\displaystyle {\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}}
3.2 Geometric optics and image formation
1 S 1 + 1 S 2 = 1 f {\displaystyle {\frac {1}{S_{1}}}+{\frac {1}{S_{2}}}={\frac {1}{f}}} relates the focal length f of the lens, the image distance S1 , and the object distance S2 . The figure depicts the situation for which (S1 , S2 , f) are all positive: (1)The lens is converging (convex ); (2) The real image is to the right of the lens; and (3) the object is to the left of the lens. If the lens is diverging (concave ), then f < 0. If the image is to the left of the lens (virtual image ), then S2 < 0 .