Question

A charge q₁ = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q₂ = -2.00 nC is placed on the positive x-axis at x = 4.00 cm. (a) If a third charge q₃ = +6.00 nC is now placed at the point x = 4.00 cm, y = 3.00 cm, find the x- and y-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

Short answer

(a) The x and y components of the total force exerted on q₃are $-3\times {10}^{-5}\mathrm{N}$and respectively $-6.5\times {10}^{-5}\mathrm{N}$.

(b) The magnitude of this force is $7.15\times {10}^{-9}\mathrm{N}$and the direction is $245°$from +x axis.

Step-by-step solution

Orientation of charges

The charges are aligned as given in the figure:

According to superposition principle:

$\begin{array}{r}{\mathbf{F}}_{\mathbf{x}}\mathbf{=}{\mathbf{F}}_{\mathbf{21}}\mathbf{-}{\mathbf{F}}_{\mathbf{31}}\mathbf{cos}\mathbf{}\mathbf{\theta }\\ {\mathbf{F}}_{\mathbf{y}}\mathbf{=}\mathbf{-}{\mathbf{F}}_{\mathbf{13}}\mathbf{sin}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Components of force

x direction of force is given by

$\begin{array}{r}{F}_x=\frac{k{q}_1{q}_2}{r_1^2}+\frac{k{q}_1{q}_3}{r_2^2}\times \frac{x}{r_2}\\ =\frac{9\times {10}^9\times 5\times {10}^{-9}\times 2\times {10}^{-9}}{(0.04{)}^2}-\frac{9\times {10}^9\times 5\times {10}^{-9}\times 6\times {10}^{-9}\times 0.04}{{\left[(0.04{)}^2+(0.03{)}^2\right]}^{3/2}}\\ =-3\times {10}^{-5}\mathrm{N}\end{array}$

y direction of force component is given by:

$\begin{array}{r}{F}_y=\frac{k{q}_1{q}_3}{r_2^2}\times \frac{y}{r_2}\\ =\frac{9\times {10}^9\times 5\times {10}^{-9}\times 6\times {10}^{-9}\times 0.03}{{\left[(0.04{)}^2+(0.03{)}^2\right]}^{3/2}}\\ =-6.5\times {10}^{-5}\mathrm{N}\end{array}$

Therefore, the x and y components of the total force exerted on q₃are $-3\times {10}^{-5}N$and $-6.5\times {10}^{-5}\mathrm{N}$respectively.

Magnitude and direction of force

Magnitude of f is given by

$\begin{array}{r}F=\sqrt{F_x^2+F_y^2}\\ =\sqrt{{\left(3\times {10}^{-5}\right)}^2+{\left(6.5\times {10}^{-5}\right)}^2}\\ =7.15\times {10}^{-9}\mathrm{N}\end{array}$

The angle is given by

$\alpha =\arctan \left(\frac{6.5}{3}\right)={65}^{\circ }$

But is in 4^th quadrant so add 180

$\alpha =65+180={245}^{\circ }$

Therefore, the magnitude of this force is $7.15\times {10}^{-9}N$and the direction is $245°$from +x axis.