Question

The charges and coordinates of two charged particles held fixed in an x-yplane are ${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{m}\mathit{C}\mathbf{,}{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathit{c}\mathit{m}\mathbf{,}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathit{c}\mathit{m}\mathbf{,}$and ${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{C}\mathbf{,}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\hspace{0.33em}}\mathit{c}\mathit{m}\mathbf{.}$Find the (a) magnitude and (b) direction of the electrostatic force on particle 2 due to particle 1. At what (c) xand (d) ycoordinates should a third particle of charge q₃=+4.0 mC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

Short answer

1. The magnitude of the electrostatic force on particle 2 due to particle 1 is 35N.
2. The direction of the electrostatic force on particle 2 due to particle 1 is -10.$3^{\circ }$
3. The x-coordinate of the third charge for net force on it to be zero is -8.4 cm
4. The y-coordinate of the third charge for net force on it to be zero is 2.7cm

Step-by-step solution

The given data

Charge $q_1=+3.0\mu C,locatedat{x}_1=3.5cm,y_1=0.50cm$

Charge $q_2=-4.0mC,locatedat{x}_2=-2.0cm,y_2=1.50cm$

The net force on the third particle of charge q₃=+4.0mC is zero.

Understanding the concept of Coulomb’s law

First, the separation between two particles is found by taking the formula of the distance between two points in a 2-dimensional structure. Using this in the given formula of Coulomb force, the magnitude is found. Using the angle formula, the direction of the force is found. Again using the same Coulomb force, the value of x and y-coordinates of the third particle is derived.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_1=k\left(\frac{\left|q_1\right|\left|q_2\right|}{r^2}\right)$ (1)

The distance between two points in the x-y plane, role="math" localid="1662621129802" $r_{12}=\sqrt{\left(\right(x_2-x_1{)}^2+(y_2-y_1{)}^2)}$ (2)

The direction of force F₂₁ directed towards charge 1 making an angle with the x-axis,

$\theta =ta{n}^{-1}\frac{(y_2-y_1)}{(x_2-x_1)}\left(3\right)$

a) Calculation of magnitude of the force on particle 2 due to 1

The distance between particles 1 and 2 is calculated by substituting the given values in equation (2) as:

$$\begin{gathered} r_{12}=\sqrt{{\left(-0.020m-0.035m\right)}^2+{(0.015m-0.005m)}^2} \\ =0.056m \end{gathered}$$

The magnitude of the force exerted by particle 1 on particle 2 is given using equation (1) as follows:

$$\begin{gathered} F_{12}=\frac{\left(9.0\times {10}^9{\displaystyle \frac{N.m^2}{C^2}}\right)\left(3.0\times {10}^{-6}C\right)\left(4.0\times {10}^{-6}C\right)}{{\left(0.056m\right)}^2} \\ =35N \end{gathered}$$

Hence, the value of the magnitude of the force is 35N

b) Calculation of direction of the force on particle 2 due to 1

The vector F₂₁ is directed towards charge 1 and makes an angle with the +x-axis, which is given using equation (3) as follows:

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{1.5\mathrm{cm}-0.5\mathrm{cm}}{-20\mathrm{cm}-3.5\mathrm{cm}}\right)\\ =-10.3^0\end{array}$

Hence, the direction of the force with the positive x-axis is -10.$3^{\circ }$

c) Calculation of x-coordinate of the third particle

Let the third charge be located at (x₃, y₃,), a distance r from q₂. We note that q₁, q₂, and q₃ must be collinear; otherwise, an equilibrium position for any one of them would be impossible to find. Furthermore, we cannot place q₃ it on the same side as q₂ where we also find q₁, since in that region both forces (exerted on q₂ by q₃ and q₁) would be in the same direction (since q₂ is attracted to both of them). Thus, in terms of the angle found in part (a), we have

$x_3=x_2-rcos\theta ................\left(4\right)$

And

$y_3=y_2-rsin\theta ....................\left(5\right)$

(which means $y_3>y_2$since $\theta$ is negative). The magnitude of the force exerted on q₂ by q₃ is $F_{23}=k\frac{\left|q_2\right|\left|q_3\right|}{r^2}$, which must equal that of the force exerted on it by q₁ (found in part (a)). Thus, using the required values and equation (1) for the condition of the net force on charge 3 is zero, we get that

$$\begin{gathered} k\frac{\left|q_2\right|\left|q_3\right|}{r^2}=k\frac{\left|q_1\right|\left|q_2\right|}{{r^2}_{12}} \\ r=r_{12}\sqrt{\frac{q_3}{q_1}} \\ r=(0.056\hspace{0.33em}m)\sqrt{\frac{+4mC}{+3\mu C}} \\ r=0.0645\hspace{0.33em}m\frac{100cm}{1m} \\ r=6.45cm \end{gathered}$$

Consequently, the x-coordinate of the third particle is found by using equation (4) as:

$$\begin{gathered} X_3=-2.0\mathrm{cm}-(6.45\mathrm{cm})c\mathrm{os}(-{10}^0) \\ =-8.4\mathrm{cm} \end{gathered}$$

Hence, the value of the x-coordinated is -8.4cm

d) Calculation of y-coordinate of the third particle

Similarly, substituting the value of r in equation (5) in part (a) calculations, we can get the y-coordinate of the third particle as:

$$\begin{gathered} y_3=1.5\mathrm{cm}-(6.45\mathrm{cm})sin(-{10}^0) \\ =2.7\mathrm{cm} \end{gathered}$$

Hence, the value of the y-coordinate of the third particle is 2.7cm.