Respuesta :
Answer:
[tex]F=6\times 10^{-7}\ N[/tex]
Explanation:
Given:
- quantity of point charge, [tex]q=-4\times 10^{-9}\ C[/tex]
- radial distance from the linear charge, [tex]r=0.09\ m[/tex]
- linear charge density, [tex]\lambda=3\times 10^{-9}\ C.m^{-1}[/tex]
We know that the electric field by the linear charge is given as:
[tex]E=\frac{\lambda}{2\pi.\epsilon_0.r}[/tex]
[tex]E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}[/tex]
[tex]E=150\ N.C^{-1}[/tex]
Now the force on the given charge can be given as:
[tex]F=E.q[/tex]
[tex]F=150\times 4\times 10^{-9}[/tex]
[tex]F=6\times 10^{-7}\ N[/tex]
