电场与电势
$$ k = \frac{1}{4\pi\varepsilon_0} = \frac{\mu_0}{4\pi} $$
$$ F_{12} = k \frac{q_1q_2}{r^2}\overrightharpoon{e_{12}} $$
$$ E = \frac{F}{q_0} $$
$$ \overrightharpoon{E} = - \nabla U $$
$$ U = \int_P^\infty\overrightharpoon{E}\cdot\overrightharpoon{dl} $$
2024/11/4...大约 1 分钟