Question

Three charges, q1=+q, q2=-q, and q3=+q, are at the vertices of an equilateral triangle, as shown in the figure.
***Part A***
Rank the three charges in order of decreasing magnitude of the electric force they experience. Rank charges from largest to smallest magnitude of the electric force they experience. To rank items as equivalent, overlap them.
***Part B***
Give the direction angle, theta, or the net force experienced by charge 1. Note that theta is measured counterclockwise from the positive x axis. Express your answer as an integer.
***Part C***
Repeat part B for charge 2.
Express your answer as an integer.
***Part D***
Repeat part B for charge 3.
Express your answer as an integer.

Part A Review Three charges, g and g triangle, as shown in the figure (Figure 1) Rank the three charges in order of decreasing magnitude of the electric force they Rank charges from largest to smallest magnitude of the electric force they experience. To rank items as equivalent, overlap them. +qg, are at the vertices of an equilateral Reset Holp Larges i The comect ranking cannot be determined Figure Submit Recuest Answer Part B Give ta direcson angle·of the net electric force experienced by charge 1 Note that θ is measured counterdockwise from the positive z axis. Express your answer as an integer degrees Submit

Answer 1

Electrostatic force between two charged particles having charge q and Q, and separated by a distance r is given by

                                                      $F = \frac{1}{4 \pi \epsilon_{0}} \frac{qQ}{r^{2}}$

Where,

                                                      $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \,\, Nm^{2}/C^{2}$

Consider, the length of the sides of equilateral triangle is a.

And, $F_{ij}$ is force on $i_{th}$ charge particle due to the $j_{th}$ particle.

9-19 Aco > Paret net T60° TE= doctor what are 1 Feb - 1926 Foto 1 - (negative sign is denoting the direction) Similarly, ET - 1) = 16.1= 16 ) = 32

From the diagram, the net force on charge $q_{1}$ is

                                                      $\vec{F}_{1_{net}} = \vec{F}_{13} + \vec{F}_{12}$

                                                                $=F_{13}\, cos \,60^{\circ} \, \hat{i} + F_{13}\, sin \,60^{\circ} \, \hat{j}+F_{12}\, cos \,60^{\circ} \, \hat{i} -F_{12}\, sin \,60^{\circ} \, \hat{j}$

                                                                $=2F_{13}\, cos \,60^{\circ} \, \hat{i} = \frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \,\, \hat{i}$

Therefore,

                                       $\left | \vec{F}_{1_{net}} \right | = \frac{q^{2}}{4 \pi \epsilon_{0}a^{2}} =\frac{2q^{2}}{8 \pi \epsilon_{0}a^{2}}$

the net force on charge $q_{2}$ is

                                                      $\vec{F}_{2_{net}} = \vec{F}_{23} + \vec{F}_{21}$

                                                                $=F_{23} \, (-\hat{j}) +F_{21}\, sin \,60^{\circ} \, \hat{i} + F_{21}\, cos \,60^{\circ} \, (-\hat{j})$

                                                                $=\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \, (-\hat{j}) +\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \times \frac{\sqrt{3}}{2} \, \hat{i} + \frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \times \frac{1}{2} \, (-\hat{j})$

                                                               $=-\frac{3q^{2}}{8 \pi \epsilon_{0} a^{2}} \, \hat{j} +\frac{\sqrt{3}q^{2}}{8 \pi \epsilon_{0} a^{2}} \, \hat{i}$

Therefore,

                                      $\left | \vec{F}_{2_{net}} \right | = \frac{q^{2}}{8 \pi \epsilon_{0}a^{2}}\left ( \sqrt{3^{2} + (\sqrt{3})^{2}} \right ) = \frac{3.464\,q^{2}}{8 \pi \epsilon_{0}a^{2}}$

                                      

the net force on charge $q_{3}$ is

                                                      $\vec{F}_{3_{net}} = \vec{F}_{32} + \vec{F}_{31}$

                                                                $=F_{32} \, \hat{i}+F_{31}\, sin \,60^{\circ} \, (-\hat{i}) + F_{31}\, cos \,60^{\circ} \, (-\hat{j})$

                                                                $=\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \, \hat{i} +\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \times \frac{\sqrt{3}}{2} \, (-\hat{i}) + \frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \times \frac{1}{2} \, (-\hat{j})$

                                                               $=-\frac{q^{2}}{8 \pi \epsilon_{0} a^{2}} \, \hat{j} +\frac{(2-\sqrt{3})q^{2}}{8 \pi \epsilon_{0} a^{2}} \, \hat{i}$

Therefore,

                                      $\left | \vec{F}_{3_{net}} \right | = \frac{q^{2}}{8 \pi \epsilon_{0}a^{2}}\left ( \sqrt{(-1)^{2} + (2-\sqrt{3})^{2}} \right ) = \frac{1.035\,q^{2}}{8 \pi \epsilon_{0}a^{2}}$

Part A:

Hence,              $\left | \vec{F}_{2_{net}} \right | > \left | \vec{F}_{1_{net}} \right | > \left | \vec{F}_{3_{net}} \right |$

Part B:

The angle made by $\vec{F}_{1_{net}}$ with positive x-axis in counterclockwise direction is 0^o as the net force is in x-direction.

                                                    $\left ( \vec{F}_{1_{net}} = \frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \,\, \hat{i} \right )$

Part C:

The angle made by $\vec{F}_{2_{net}}$ with positive x-axis is

                                                   $\alpha = tan^{-1}\left ( \frac{-3}{\sqrt{3}} \right ) = -60^{\circ}$      (in fourth quadrant)

Hence, The angle made by $\vec{F}_{2_{net}}$ with positive x-axis in counterclockwise direction is (360-60)=300^o .

Part D:

The angle made by $\vec{F}_{3_{net}}$ with positive x-axis is

                                                   $\theta = tan^{-1}\left ( \frac{-1}{2-\sqrt{3}} \right ) = -75^{\circ}$      (in fourth quadrant)

Hence, The angle made by $\vec{F}_{2_{net}}$ with positive x-axis in counterclockwise direction is (360-75)=285^o .

For any doubt please comment.