Review: Capacitors ΔV = Q/C (finding equivalent capacitor: Q, ΔV don't change)
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Key Points
| Current Density | $$ { \vec J = \, n \, e \, \vec v_d \, = \, n \, e \, \left( { {e \tau} \over m} \vec E \right) \equiv \sigma \vec E \equiv {1 \over \rho} \vec E \, \,\, [Am^{-2}] }$$ |
Ohm's Law | |
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| $$ I \equiv \iint \vec J \cdot d \vec A = {1 \over \rho} \iint \vec E \cdot d \vec A = {1 \over \rho} \iint \nabla V \cdot d \vec A \, \\ \rightarrow I = {A \over {\rho L}} \Delta V \equiv {{\Delta V} \over R} $$ | |
| Assumptions | 1. ρ independent of space (x,y,x) & E-field (i.e. drift velocity is linearly depends on electric field. (i.e. τ is a constant)) |
| 2. E-field is a linear function of space | |
Kirchoff's Laws | |
| 1: Energy Conservation | $$ \Delta V_{closed \, loop} = \sum \Delta V_i = 0 \, $$ |
| 2: Charge Conservation | $$ \sum I_{in} - \sum I_{out} = 0 $$ |
- Nonuniform Charge Distribution Along surface of wire
→ longitudinal E-Field in wire
→ current flow in wire - Battery is the source of everything: ℰ = W/q
→ ΔVbat = (1 - δ) ℰ (internal resistance)
→ Δ Vwire
→ Ewire = Δ Vwire/L
→ I = σ A Ewire = A/ρ Ewire
Magnitude of current is related to both battery's ℰ and wire's resistance: I = ΔVbat/Rwire