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# Physics Equations: E&M

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Part 2 of the equation posts: E&M. Again, if you see any mistakes or have a few equations to add, make sure to utilize the comment section! I'll add it in right away.

Electrostatics

$F_e=\frac{kq_1q_2}{r^{2}}$

E= Fe/q = kq/r

λ = Q/L

ρ = Q/V

σ = Q/A

Electric potential

Ue = kq1q2/r

F = -dU/dl

V = k ∑ qi/ri = W/q

∆V= Vb - Va = ab E dl = ∆U/q

Gauss's Law:

$\int\limits_{closed surface} {\vec E \bullet d\vec A} = {{{Q_{enc}}} \over {{\varepsilon _0}}}$

Conductors

Esurface= $\frac{\sigma}{\epsilon_{0} }$

Vinside = $\frac{Q}{4\pi \epsilon _{0}R}$

Einside = $0$

Capacitor

C=Q/V = $\frac{\epsilon _{0} A}{d}$

Uc =$\frac{1}{2}CV^{2}$

Ue = field energy density = $\frac{U}{volume}$

Energy = V/d

C= $\frac{\kappa \epsilon _{0}A }{d}$ w/ $\kappa$ = Dielectric constant

Circuits

I = $\frac{\Delta Q}{\Delta t}$

I = V/R

I = NqVdA = NeVdA

Current density (J) = NqVd = I/A

charging up: w = I2V

charging down: U = $\frac{1}{2}$ CV2

$\tau$ = RC

5 $\tau$ = 99% charged/discharged

Resistance

R= $\frac{\rho L}{A}$

E= $\frac{V}{L}$

P=IV

E= $\rho$ J

W= qv

Series Circuit

Ceq = $\frac{Q}{V}$

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + ...$

I= constant

V=IR

Req= R1+R2+...

Q=CV

Parallel Circuit

Ceq = C1 + C2 + ...

I = V/R

V= constant

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ...$

Q=CV

Batteries:

Videal = $\Delta$ V = $\varepsilon$mf = VT

V with resistance = $\varepsilon$ Iri = VT

Pbattery = W/t = $\frac{\Delta \varepsilon}{t}$= I$\varepsilon$

Pexternal resistor = I2R

Pinternal resistor = I2ri

Magnetism

Gauss's Law: $\phi = \oint B \cdot$ dA = O

Amphere's Law: $\oint B \cdot dl = \mu_{0}$ Ipen

...for a wire of radius R, B = $\frac{\mu_{0} I }{2 \pi R^{2}}$

Biot-Savart law: dB = $\frac{\mu_{0} I}{4 \pi r^{2}}$ (dl X r)

...for a loop of wire, B = $\frac{\mu_{0} I}{2r}$

Faraday's Law: $\varepsilon = \frac{d \phi_{B}}{dt}$

= $\frac{-d}{dt} \oint_{open surface} B \cdot dA$

= $\int E \cdot dl$

solenoid: B = $\frac{mu_{0} I N}{L}$

toroid: B = $\frac{mu_{0} I N}{2 \pi r}$

Mag. moment ($\mu$) =NIA =NI$\pi$R2

Mag. torque= $\mu_{0} X B = \mu_{0} B sin\theta$

...sorry that took so long to post up, jeez that code takes a while to type up ^-^ Feel free to add/correct in the comments section!

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