Talk:QB/a19ElectricPotentialField KE PE

$Δ V_{A B} = V_{A} - V_{B} = - \int_{A}^{B} \vec{E} \cdot d \vec{ℓ}$ = electric potential

$\vec{E} = - \frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j} - \frac{\partial V}{\partial z} \hat{k} = - \vec{\nabla} V$

$q Δ V$ = change in potential energy (or simply $U = q V$ )

$P o w e r = \frac{Δ U}{Δ t} = \frac{Δ q}{Δ t} V = I V = e \frac{Δ N}{Δ t}$

Electron (proton) mass = 9.11×10⁻³¹kg (1.67× 10⁻²⁷kg). Elementary charge = e = 1.602×10⁻¹⁹C.

$K = \frac{1}{2} m v^{2}$ =kinetic energy. 1 eV = 1.602×10⁻¹⁹J

$V (r) = k \frac{q}{r}$ near isolated point charge

Many charges: $V_{P} = k \sum_{1}^{N} \frac{q_{i}}{r_{i}} \to k \int \frac{d q}{r}$ .

The alpha-particle is made up of two protons and two neutrons.

$Q = C V$ defines capacitance.

$C = ε_{0} \frac{A}{d}$ where A is area and d<<A^1/2 is gap length of parallel plate capacitor

$Series : \frac{1}{C_{S}} = \sum \frac{1}{C_{i}} .$ $Parallel: C_{P} = \sum C_{i} .$

$u = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{1}{2 C} Q^{2}$ = stored energy

$u_{E} = \frac{1}{2} ε_{0} E^{2}$ = energy density

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