Force due to point charges is given by Coulomb's law:

$\overline{){\mathbf{F}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{1}}}{{\mathbf{r}}^{\mathbf{2}}}}$

2D vector magnitude:

$\overline{)\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}{\mathbf{=}}\sqrt{{{\mathit{A}}_{\mathit{x}}}^{\mathbf{2}}\mathbf{+}{{\mathit{A}}_{\mathit{y}}}^{\mathbf{2}}}}$

2D vector direction:

$\overline{){\mathbf{tan}}{\mathit{\theta}}{\mathbf{=}}\frac{{\mathit{A}}_{\mathit{y}}}{{\mathit{A}}_{\mathit{x}}}}$

Resolving a vector into its components:

$\overline{)\begin{array}{rcl}{\mathit{A}}_{\mathit{x}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}\mathbf{cos}\mathbf{}\mathit{\theta}\\ {\mathit{A}}_{\mathit{y}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\theta}\end{array}}$

**(1)**

r_{13}^{2} = (0.04^{2} + 0.302^{2})

q_{1} = 5.02 nC (10^{-9C}/1nC) = 5.02 × 10^{-9} C

q_{3} = 6.03 nC (10^{-9C}/1nC) = 6.03 × 10^{-9} C

A charge 5.02 nC is placed at the origin of an xy-coordinate system, and a charge -2.05 nC is placed on the positive x-axis at x = 4.00cm. A third particle, of charge 6.03 nC is now placed at the point x = 4.00 cm , y = 3.02 cm .

1. Find the *x*-component of the total force exerted on the third charge by the other two.

2. Find the *y*-component of the total force exerted on the third charge by the other two.

3. Find the magnitude of the total force acting on the third charge.

4. Find the direction of the total force acting on the third charge. ( *r**a**d**i**a**n**s* between *F*? and +*x*-axis )

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