Oblicz odległość , z jakiej dwa ładunki + 1µC będą się odpychać siłą 1 N . 1 µC = 10 – 6 C k = 9 ∙10 9 N∙m2 / kg2

Odpowiedź :

[tex]dane:\\q_1 = q_2 = q = +1 \ \mu C = +10^{-6} \ C\\F = 1 \ N\\k = 9\cdot10^{9}\frac{Nm^{2}}{kg^{2}}\\szukane:\\r = ?\\\\\\F = k\cdot\frac{q_1q_2}{r^{2}}\\\\q_1 = q_2 = q\\\\F = k\cdot\frac{q^{2}}{r^{2}} \ \ |\cdot r^{2}\\\\F\cdot r^{2} = k\cdot q^{2} \ \ /:F\\\\r^{2} = k\cdot\frac{ q^{2}}{F}\\\\r = \sqrt{k\cdot\frac{q^{2}}{F}}\\\\r = \sqrt{9\cdot10^{9}\frac{Nm^{2}}{C^{2}}\cdot\frac{(10^{-6}C)^{2}}{1 \ N} }=10^{-6}\cdot\sqrt{90\cdot10^{8}} \ m\approx10^{-6}\cdot9,4868\cdot10^{4} \ m[/tex]

[tex]r\approx9,5\cdot10^{-2} \ m \approx9,5 \ cm[/tex]