Расчет: 2pfpOG - 1
φ =
| k* q0 |
| / ( ε* r) |
|
φ =
$$\phi$$ = $$\frac{k\cdot q0}{\varepsilon\cdot r}$$
| k* q0 |
| / ( ε* r) |
φ =
$$\phi$$ = $$\frac{k\cdot q0}{r\cdot \varepsilon}$$
| k* q0 |
| / r/ ε |
| k* q0 |
| / ( ε* r) |
|
|
| k* q0 |
| / ( ε* r) |
| k* q0 |
| / r/ ε |