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Physics II - B05

TABLE OF CONTENTS

Work (Work done by the field)

\[\displaystyle W_{field} = \int \vec F. d\vec s = \int F.ds.cos\theta = \int_{line} q.\vec E. d\vec s\]

Work - KE theorem of a particle

\[W_{field} = \Delta K = K_f -K_i\]

System = a charge + the field : isolated

\[\Delta K - W_{field} = 0 \, (W_{field} = \Delta U)\] \[\Delta K + \Delta U = 0 \to \Delta E = 0\] \[\displaystyle \Delta U = - \int_{initial}^{final} q.\vec E. d\vec s = U_f - U_i = q.\Delta V\] \[\to \Delta V = \frac{\Delta U}{q} = V_f - V_i\]

Therefore: \(\displaystyle Choose \, V_i = 0 \to V_f = -\int_i^f \vec E.d \vec s\)

[energy] = 1eV = $1,6.10^{-19} (J)$ [electric field] = $\frac{V}{m}$ = $\frac{N}{C}$

Charged Particle in a Uniform Field

\[\displaystyle \Delta V = -\int_i^f \vec E.d\vec s = - \int_i^f E.ds = -E \int_i^f ds = -E.d < 0 \to V_f < V_i\]

Equipotential Surface (Mặt đẳng thế)

Potential and Point Charges

\[\displaystyle \Delta V = -\int_A^B \vec E.d\vec s = -\int_{circular}\vec E. d\vec s - \int_{radial}\vec E. d\vec s\] \[\displaystyle \int_{circular}\vec E. d\vec s = 0\] \[\displaystyle = -\int_{r_A}^{r_B}\frac{kQ.dr}{r^2}\] \[= kQ.\left(\frac{1}{r_B} - \frac{1}{r_A}\right) = V_B - V_A\]

Notes

  • Vecto $\bot$ với mặt cong $\Leftrightarrow$ Vecto $\bot$ tiếp tuyến của mặt cong tại điểm đó.
  • Orthonomal: trực chuẩn
  • Potential energy of the system:
    • Charge
    • Field

System of charges

\(\displaystyle \Delta V = -\int E_{net}.d \vec s = - \int(\sum \vec E).d \vec s\)

\[\displaystyle = -\sum \int\vec E.d\vec s = \sum \Delta V\]

Choose $V_{\infty} = 0:$ \(V = \sum\frac{k.Q_i}{r_i}\)

Assignments

Other references