Equations Flashcards
Coulomb Force
- F_12 = (q_1 q_2)/(4πϵ) x (r2 - r1)/r^3
Electric Field
E = (q/4πϵ) x (r - rq)/(r^3)
Electric Dipole Moment
p = qd
Torque on electric dipole
t = p x E
Electric Flux through flat surface
Φ = E · A
Electric Flux INtegral
Φ = ∯E· dS
Gauss’s Law
∯E· dS = Q/ϵ
Work Done and Potential Energy
W = ∫F · dl = -ΔU
Potential Energy between 2 point charges
U = qq/4πϵr
Electric Potential of a point charge
V = q/4πϵr
Capacitance
C = Q/V = ϵA/d
Energy stored in a capacitor
U = QV/2
Energy density of an electric field
u = (ϵE^2)/2
Current Density Vector
J = I/A
Lorentz Force
F = q(E + v x B)
Magnetic Flux
Φ = ∬B · dS
Solenoid Condition
∯B · dS = 0
Force on a current carrying wire
dF = IdL x B
Magnetic Dipole Moment
μ = IA n
Torque on a current loop
t = μ x B
Biot Savart Law
B = (μq/4π) (v(t) x (r - r(t))) / |r -r(t)|^3
Magnetic Field of a straight wire
B = (Iμ0/2πR) φ
Ampere’s Law
∮B · dL = μ0 I
Magnetic Field of a solenoid
- Outside = 0
- Inside = μ0 I n z
Faradays Law of induction
ϵ = -(dΦ)/(dt)
Self Inductance
L = NΦ/I
Flux Linkage
NΦ
Emf of self inductor
ϵ = -L (dI/dt)
Energy stored in a inductor
U = (LI^2)/2
Impedance of a Resistor
Z = R
Impedance of a capacitor
Z = 1/(jwC)
Impedance of an inductor
Z = jwL
Electric Field from Potential
- E = - gradient of V
Capacitor Discharging Equations
- A = A0 x e^(-t/RC)
- A is Q, V and I
- I is same for charging and discharging
Capacitor Charging Equations
- A = A0 (1 - e^(-t/RC))
- A is Q and V
- I is same for charging and discharging
Faradays Law of Induction Relation with electric fields
ϵ = -dϕ/dt = ∮E·dl
Potential (difference) from Electric Field equation
Va - Vb = ∫E ·dl
