Answer:
[tex]\vec{F} = -0.34\^x - 0.22\^y\\|\vec{F}| = -0.41~N[/tex]
Explanation:
The electric force between two point charges can be calculated by Coulomb's Law:
[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]
We have to calculate the distance between two points; (0,0) and (0.3 m, 0.2 m).
[tex]r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0.3)^2 + (0.2)^2} = 0.36~m[/tex]
Now we can apply Coulomb's Law
[tex]F = \frac{1}{4\pi\epsilon_0}\frac{(3\times 10^{-6})(-2\times 10^{-6})}{(0.36)^2} = -0.41~N[/tex]
The minus sign in front of the force means that the force is attractive.
The direction of the force can be calculated as follows:
[tex]F_x = F\cos(\theta)\\F_y = F\sin(\theta)[/tex]
where θ is the angle between F and the x-axis. This angle can be calculated by the triangle with edges 0.3 m, 0.2 m, and 0.36 m.
So, sin(θ) = 0.2/0.36 = 0.55 and cos(θ) = 0.3/0.36 = 0.83.
Finally,
[tex]F_x = -0.41 \times 0.83 = -0.34~N\\F_y = -0.41 \times 0.55 = -0.22~N[/tex]
