Particle A has a charge of 33.5 nC, and particle B has a charge of −50.3 nC. The attractive force between them has a magnitude of 3.30 ✕ 10−4 N. How far apart are the particles?

Respuesta :

Answer:

Distance between both the charges will be equal to [tex]r=22.59\times 10^7m[/tex]

Explanation:

We have first charge [tex]q_1=33.5nC=33\times 10^{-9}C[/tex]

And charge [tex]q_2=-50.3nC=-50.3\times 10^{-9}C[/tex]

Force between theses charges are equal to [tex]F=3.30\times 10^{-4}N[/tex]

We have to find the distance between two charges

According to coulomb's law force between two charges is equal to [tex]F=\frac{Kq_1q_2}{r^2}[/tex]

So [tex]3.3\times 10^{-4}=\frac{33.5\times 10^{-9}\times 50.3\times 10^{-9}}{r^2}[/tex][tex]r=22.59\times 10^7m[/tex]

So distance between both the charges will be equal to [tex]r=22.59\times 10^7m[/tex]