Fizyka zadanie dwa jednakowe pubktowe ladunki q = 10...

[tex]dane:\\q_1 = q_2 = q = 10^{-4} \ C\\F = 90 \ N\\k = 9\cdot10^{9}\frac{Nm^{2}}{C^{2}}\\szukane:\\r = ?\\\\Rozwiazanie\\\\F = k\cdot\frac{q_1\cdot q_2}{r^{2}}\\\\q_1 = q_2 = q\\\\wiec:\\\\F = k\cdot\frac{q^{2}}{r^{2}} \ \ |\cdot r^{2}\\\\r^{2}\cdot F = k\cdot q^{2} \ \ |:F\\\\r^{2} = \frac{k\cdot q^{2}}{F}\\\\r = \sqrt{\frac{k\cdot q^{2}}{F}}\\\\r = \sqrt{\frac{9\cdot10^{9}\frac{Nm^{2}}{C^{2}}\cdot(10^{-4} \ C)^{2}}{90 \ N}}\\\\r = \sqrt\frac{9\cdot10^{9}\cdot10^{-8}}{90}} \ m\\\\\underline{r = 1 \ m}[/tex]

Marokesss Marokesss · Answer 2 · 2021-10-17T13:37:25+02:00

Odpowiedź:

r = 1 m

Wyjaśnienie:

q - ładunek , q = [tex]10^{-4} C[/tex]

F - siła , F = 90 N

k - stała , [tex]k =[/tex] [tex]9*10^{9} \frac{Nm^{2} }{C^{2} }[/tex]

r - odległość , r = ?

[tex]F = k*\frac{q*q}{r^{2} }[/tex] /*[tex]r^{2}[/tex]

F * [tex]r^{2}[/tex] = k * [tex]q^{2}[/tex] /:F

[tex]r^{2}[/tex] = [tex]\frac{k*q^{2} }{F}[/tex] = k * [tex]\frac{q^{2} }{F}[/tex] = [tex]9*10^{9} \frac{Nm^{2} }{C^{2} }[/tex] * [tex]\frac{(10^{-4}C)^{2} }{90 N}[/tex] = [tex]9*10^{9} \frac{Nm^{2} }{C^{2} }[/tex] * [tex]\frac{10^{-8}C^{2} }{90N}[/tex]

[tex]r^{2}[/tex] = 90 * [tex]10^{8}[/tex] * [tex]\frac{10^{-8} }{90}[/tex] [tex]m^{2}[/tex] = 1 [tex]m^{2}[/tex]

r = 1 m