Determining the Net Force Acting on Q2 at the Apex of an Equilateral Triangle
In this article, I will walk you through the process of determining the net electric force acting on a charge located at the apex of an equilateral triangle when two other charges are placed at the other corners. The article provides a step-by-step breakdown of the calculations using Coulomb's Law and the superposition principle.
Problem Restatement and Assumptions
The original problem is poorly worded, and we will assume the following to make it clear:
The sides of the equilateral triangle are (2s). The origin is at the midpoint of the base, and the base lies on the x-axis. (Q_1 Q_3 5 mu C) There was a typo in the sine and cosine values: (cos 60^{circ} 0.5) and (sin 60^{circ} 0.8) (Q_1) and (Q_3) are on the base corners, and (Q_2) sits on the y-axis at the apex.Theoretical Background
Electric forces can be calculated using Coulomb's Law, which states that the force between two point charges is given by:
[ F kfrac{q_1 q_2}{r^2} ]
Where:
(k) is Coulomb's constant, (k 9 times 10^9 text{ N m}^2/text{C}^2) (q_1) and (q_2) are the magnitudes of the charges (r) is the distance between the chargesThe superposition principle aids in finding the net force by stating that the net force on a particle is the vector sum of the individual forces acting on it.
Problem Setup
Let's define the positions of the charges:
(Q_1 Q_3 5 mu C).' (Q_2 3 mu C). The base of the equilateral triangle lies on the x-axis, and the triangle’s side length is (2s). (Q_2) is located at a height of (s sqrt{3}) above the x-axis.This configuration allows us to use geometric principles and Coulomb's Law to determine the forces.
Calculations
We need to determine the forces exerted by (Q_1) and (Q_3) on (Q_2).
Forces on (Q_2) due to (Q_1)
The position vector from (Q_1) to (Q_2) is:
[ vec{r_{12}} s hat{i} - s sqrt{3} hat{j} ]
The magnitude of (vec{r_{12}}) is:
[ r_{12} |vec{r_{12}}| sqrt{(s)^2 (s sqrt{3})^2} 2s ]
The unit vector (hat{r_{12}}) is:
[ hat{r_{12}} frac{vec{r_{12}}}{r_{12}} frac{s hat{i} - s sqrt{3} hat{j}}{2s} frac{1}{2} hat{i} - frac{sqrt{3}}{2} hat{j} ]
The force vector (vec{F_{12}}) is:
[ vec{F_{12}} k frac{Q_1 Q_2}{r_{12}^2} hat{r_{12}} 9 times 10^9 frac{(5 times 10^{-6})(3 times 10^{-6})}{(2s)^2} left( frac{1}{2} hat{i} - frac{sqrt{3}}{2} hat{j} right) ]
[ vec{F_{12}} frac{0.03}{(2s)^2} left( frac{1}{2} hat{i} - frac{sqrt{3}}{2} hat{j} right) text{N/C} ]
[ vec{F_{12}} frac{0.015}{s^2} hat{i} - frac{0.015 sqrt{3}}{s^2} hat{j} text{N/C} ]
Forces on (Q_2) due to (Q_3)
The position vector from (Q_3) to (Q_2) is:
[ vec{r_{32}} -s hat{i} - s sqrt{3} hat{j} ]
The magnitude of (vec{r_{32}}) is also (2s), and the unit vector (hat{r_{32}}) is the same as (hat{r_{12}}) but with opposite sign for (hat{i}):
[ hat{r_{32}} -frac{1}{2} hat{i} - frac{sqrt{3}}{2} hat{j} ]
The force vector (vec{F_{32}}) is:
[ vec{F_{32}} k frac{Q_3 Q_2}{r_{32}^2} hat{r_{32}} 9 times 10^9 frac{(5 times 10^{-6})(3 times 10^{-6})}{(2s)^2} left( -frac{1}{2} hat{i} - frac{sqrt{3}}{2} hat{j} right) ]
[ vec{F_{32}} -frac{0.015}{s^2} hat{i} - frac{0.015 sqrt{3}}{s^2} hat{j} text{N/C} ]
Net Force
The net force (vec{F_{tot}}) is the sum of (vec{F_{12}}) and (vec{F_{32}}):
[ vec{F_{tot}} vec{F_{12}} vec{F_{32}} ]
[ vec{F_{tot}} left( frac{0.015}{s^2} hat{i} - frac{0.015 sqrt{3}}{s^2} hat{j} right) left( -frac{0.015}{s^2} hat{i} - frac{0.015 sqrt{3}}{s^2} hat{j} right) ]
[ vec{F_{tot}} -frac{0.03 sqrt{3}}{s^2} hat{j} text{N/C} ]
Thus, the net force acting on (Q_2) is directed along the negative y-axis and its magnitude is:
[ F_{tot} frac{0.03 sqrt{3}}{s^2} text{N/C} ]
Where (s frac{text{side length}}{2}).
Conclusion
This process demonstrates how to determine the net electric force acting on a charge at the apex of an equilateral triangle, given the charges at the base corners. The calculations are based on Coulomb's Law and the superposition principle.
Feel free to implement this in your assignment or use it to understand the forces in similar configurations.
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