Respuesta :
Answer:
[tex]F_{net} = 4.22 N[/tex]
Explanation:
Since charge 1 and charge 2 are positive in nature so here we will have repulsion type of force between them
It is given as
[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]
[tex]F_{12} = \frac{(9\times 10^9)(5 \mu C)(24 \mu C)}{0.23^2 + 0.69^2}\frac{-0.23\hat i + 0.69 \hat j}{\sqrt{0.23^2 + 0.69^2}}[/tex]
[tex]F_{12} = 2.81(-0.23\hat i + 0.69\hat j)[/tex]
Since charge three is a negative charge so the force between charge 1 and charge 3 is attraction type of force
[tex]F_{13} = \frac{(9\times 10^9)(5 \mu C)(5 \mu C)}{0.27^2 + 0^2} (-\hat i)[/tex]
[tex]F_{13} = 3.1(- \hat i)[/tex]
Now we will have net force on charge 1 as
[tex]F_{net} = F_{12} + F_{13}[/tex]
[tex]F_{net} = (-0.65 \hat i + 1.94 \hat j) + (-3.1 \hat i)[/tex]
[tex]F_{net} = (-3.75 \hat i + 1.94 \hat j)[/tex]
now magnitude of total force on the charge is given as
[tex]F_{net} = 4.22 N[/tex]
