Respuesta :
Answer:
[tex]F = 5.6 \times 10^{-11} N[/tex]
[tex]F_x = 4.28\times 10^{-11} N[/tex]
[tex]F_y = 3.61\times 10^{-11} N[/tex]
Explanation:
Magnitude of the force on -2e charge due to another charge -3e is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we will have
[tex]F = \frac{(9\times 10^9)(2e)(3e)}{(3.8^2 + 3.2^2)\times 10^{-18}}[/tex]
[tex]F = 5.6 \times 10^{-11} N[/tex]
now for the two components of the force we know that this force always along the line joining two charges
so we will have
[tex]tan\theta = \frac{y}{x}[/tex]
[tex]\theta = tan^{-1}\frac{3.2 nm}{3.8 nm}[/tex]
[tex]\theta = 40.1 degree[/tex]
now the two components of above force will be
[tex]F_x = Fcos\theta[/tex]
[tex]F_x = (5.6\times 10^{-11})cos40.1[/tex]
[tex]F_x = 4.28\times 10^{-11} N[/tex]
[tex]F_y = Fsin\theta[/tex]
[tex]F_y = (5.6\times 10^{-11})sin40.1[/tex]
[tex]F_y = 3.61\times 10^{-11} N[/tex]
