How to Solve Coulomb's Law Question 1: Calculating Net Force | Physics Problem

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Potatochip911
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1. The problem statement, all variables and given/known }data
The picture I drew is quite sad but it's the best I could do lol.
102ti87.png

I forgot to mention in the image that it's the magnitude of the net force we're looking for.

Homework Equations


##|F_{12}|=\frac{1}{4\pi\epsilon_0}\frac{|q_1||q_2|}{(R_{12})^2}##

The Attempt at a Solution


I've tried solving it 2 ways and both have failed. The first method I tried was to ignore directional vectors since the y components cancel out in this example.
##F_{3}=F_{13}+F_{23}=\frac{1}{4\pi\epsilon_0}\frac{|q_1||q_3|}{(R_{13})^2}+\frac{1}{4\pi\epsilon_0}\frac{|q_2||q_3|}{(R_{23})^2}## ; Then since the hypotenuse of both triangles is the same we have ##R_{13}=R_{23}##; ##R_{13}=\sqrt{x^2+d^2}## and since ##q_1=q_2## we have $$F_{3}=\frac{1}{4\pi\epsilon_0}\frac{2|q_1||q_2|}{(R_{13})^2}=\frac{1}{4\pi\epsilon_0}\frac{2|q_1||q_2|}{x^2+d^2}$$
Unfortunately this isn't the correct expression, I also tried solving it with x and y components instead of magnitudes:
##F_3=\frac{1}{4\pi\epsilon_0}[\frac{|q_1||q_2|\cos\theta(+\hat{\imath})}{(R_{13})^2}+\frac{|q_1||q_2|\sin\theta(-\hat{\jmath})}{(R_{13})^2}]+\frac{1}{4\pi\epsilon_0}[\frac{|q_2||q_3|\cos\theta(+\hat{\imath})}{(R_{23})^2}+\frac{|q_2||q_3|\sin\theta(+\hat{\jmath})}{(R_{23})^2}]## which after simplifying leads to the expression $$F_3=\frac{1}{4\pi\epsilon_0}[\frac{2|q_1||q_3|\cos\theta}{(x^2+d^2)}]$$ which also isn't the correct expression even after substituting ##\cos\theta=\frac{x}{\sqrt{x^2+d^2}}##
 

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Potatochip911 said:
1. The problem statement, all variables and given/known }data
The picture I drew is quite sad but it's the best I could do lol.
102ti87.png

I forgot to mention in the image that it's the magnitude of the net force we're looking for.

Homework Equations


##|F_{12}|=\frac{1}{4\pi\epsilon_0}\frac{|q_1||q_2|}{(R_{12})^2}##

The Attempt at a Solution


I've tried solving it 2 ways and both have failed. The first method I tried was to ignore directional vectors since the y components cancel out in this example.
##F_{3}=F_{13}+F_{23}=\frac{1}{4\pi\epsilon_0}\frac{|q_1||q_3|}{(R_{13})^2}+\frac{1}{4\pi\epsilon_0}\frac{|q_2||q_3|}{(R_{23})^2}## ; Then since the hypotenuse of both triangles is the same we have ##R_{13}=R_{23}##; ##R_{13}=\sqrt{x^2+d^2}## and since ##q_1=q_2## we have $$F_{3}=\frac{1}{4\pi\epsilon_0}\frac{2|q_1||q_2|}{(R_{13})^2}=\frac{1}{4\pi\epsilon_0}\frac{2|q_1||q_2|}{x^2+d^2}$$
That is wrong, naturally.
Potatochip911 said:
Unfortunately this isn't the correct expression, I also tried solving it with x and y components instead of magnitudes:
##F_3=\frac{1}{4\pi\epsilon_0}[\frac{|q_1||q_2|\cos\theta(+\hat{\imath})}{(R_{13})^2}+\frac{|q_1||q_2|\sin\theta(-\hat{\jmath})}{(R_{13})^2}]+\frac{1}{4\pi\epsilon_0}[\frac{|q_2||q_3|\cos\theta(+\hat{\imath})}{(R_{23})^2}+\frac{|q_2||q_3|\sin\theta(+\hat{\jmath})}{(R_{23})^2}]## which after simplifying leads to the expression $$F_3=\frac{1}{4\pi\epsilon_0}[\frac{2|q_1||q_3|\cos\theta}{(x^2+d^2)}]$$ which also isn't the correct expression even after substituting ##\cos\theta=\frac{x}{\sqrt{x^2+d^2}}##
Use the value of the charge when you calculate the Coulomb force, not the absolute value.
You want the force acting on charge 3, but you used q1 q2 in your first two terms.
By the way, where is x=0?
 
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ehild said:
That is wrong, naturally.

Use the value of the charge when you calculate the Coulomb force, not the absolute value.
You want the force acting on charge 3, but you used q1 q2 in your first two terms.
By the way, where is x=0?
Okay, I had ##q_1## and ##q_3## on my paper but I accidentally put ##q_2## in LaTeX. So placing the proper variables gives $$\frac{1}{4\pi\epsilon}[\frac{q_1q_3\cos\theta(+\hat{\imath})}{(R_{13})^2}+\frac{q_1q_3\sin\theta(-\hat{\jmath})}{(R_{13})^2}]+\frac{1}{4\pi\epsilon}[\frac{q_2q_3\cos\theta(+\hat{\imath})}{(R_{23})^2}+\frac{q_2q_3\sin\theta(+\hat{\jmath})}{(R_{23})^2}]$$ which after making the substitutions ##q_1=q_2##, ##R_{13}=R_{23}## and ##\cos\theta=\frac{x}{\sqrt{x^2+d^2}}## becomes $$\frac{1}{4\pi\epsilon}\frac{2q_1q_3\cos\theta}{(R_{13})^2}=\frac{1}{4\pi\epsilon}\frac{16e^2x}{(x^2+d^2)^{3/2}}$$ which is pretty much the same expression as before just without the absolute values. x=0 would be where the y-axis intersects the x-axis in the image.
 
ehild said:
How do you know it is wrong? It looks correct, but you can substitute the the numerical values.
In the second part of the question it asked to calculate where the force would be at a maximum and I thought the derivative was elementary enough that I wouldn't make a mistake on it so I assumed the original expression was incorrect. After looking over it again it turns out it's actually the derivative that I took incorrectly and the function ##F_3(x)## is correct.