Zero Force from 3 charges placed at the vertices of a Triangle

[haruspex]

I was using the base of the triangle as the origin.

In my original equations, I had got one of the points correct when I put it in Wolfram Alpha. I just don't see why I didn't get the other correct point.

That would be because you did not use the absolute value in the denominator.
In the original equation it was of the form $(x^2)^{\frac 32}$, which is always positive. When you simplified it to $x^3$ it was wrong for negative x.

 

[Charles Link]

That would be because you did not use the absolute value in the denominator.
In the original equation it was of the form $(x^2)^{\frac 32}$, which is always positive. When you simplified it to $x^3$ it was wrong for negative x.

Correct me if I am wrong, but I think there is only one solution above the apex of the triangle. The other solution is just below the base. You need to rewrite the equations for this second point.

 

[Charles Link]

Waiting to hear from the OP to see if he gets the correct expression per post 34. The solution for $y$, and $y$ is small, can be estimated by using $\sqrt{3}+y \approx \sqrt{3}$, and $y^2+1 \approx 1$.(Note: This estimate is somewhat poor. I think the correct answer might be what the OP mentioned in post 29 at the bottom).

 

[cwill53]

Waiting to hear from the OP to see if he gets the correct expression per post 34. The solution for $y$, and $y$ is small, can be estimated by using $\sqrt{3}+y \approx \sqrt{3}$, and $y^2+1 \approx 1$.(Note: This estimate is somewhat poor. I think the correct answer might be what the OP mentioned in post 29 at the bottom).

I tried making expressions, but when I put them in Wolfram I did not get what I should have gotten.

I treated the point below my axis P(0,y) as just a point with a positive valued y, as it really shouldn't matter as long as I keep the signs consistent with each other.
The simplifications made for the variables were:

$a_1=1m,q_1=q_2=1C,q_3=-1C$
So again,
$$\vec{E}(x,y)=\frac{1}{4\pi \varepsilon _0}\sum_{j=1}^{N}\frac{q_j\hat{r}_{oj}}{r^{2}_{oj}}=\frac{1}{4\pi \varepsilon _0}\sum_{j=1}^{N}\frac{q_j\vec{r}_{oj}}{r^{2}_{oj}}\frac{1}{\left \|\vec{r} \right \|}$$

$$\vec{E}_1(x,y)=\frac{1}{4\pi \varepsilon _0} \frac{q_1\left \langle a_1,y \right \rangle}{(\sqrt{(a_1)^2+(y)^2})^3}$$
$$\vec{E}_2(x,y)=\frac{1}{4\pi \varepsilon _0} \frac{q_1\left \langle -a_1,y \right \rangle}{(\sqrt{(a_1)^2+(y)^2})^3}$$
$$\vec{E}_3(x,y)=\frac{1}{4\pi \varepsilon _0} \frac{q_3\left \langle 0,\sqrt{3}+y \right \rangle}{(\sqrt{(\sqrt{3}+y )^2})^3}$$
$$\vec{E}(0,y)=k_{e}(\frac{2y}{(a_1^2+y^2)^{\frac{3}{2}}}+\frac{-\sqrt{3}-y}{\sqrt{(y+(\sqrt{3}))^2}^3})$$
I set this equal to 0 and solved for y but got the wrong answer.

 

[Charles Link]

There are a couple of things going on with these equations that are clumsy. One thing that is complicating things is that for the single charge, instead of using inverse square, you are complicating it with a linear term divided by a cube. For the first point, you can simply write: $\frac{1}{(y-\sqrt{3})^2}=2 \frac{y}{(y^2+1 )^{3/2}}$. If you then square everything in the algebra, you can get extraneous solutions. In addition, @haruspex pointed out why you were getting an incorrect expression when the $x$ was negative. That's why it's really simpler to use a second expression to get the second point=there are otherwise too many minus signs floating around that should be with an absolute value sign.
Note: If you change the minus in the first denominator to a plus (and y going downward is positive), I think that gives you the expression for the second point. The solution really is kind of simple.

 

[cwill53]

There are a couple of things going on with these equations that are clumsy. One thing that is complicating things is that for the single charge, instead of using inverse square, you are complicating it with a linear term divided by a cube. For the first point, you can simply write: $\frac{1}{(y-\sqrt{3})^2}=2 \frac{y}{(y^2+1 )^{3/2}}$. If you then square everything in the algebra, you can get extraneous solutions. In addition, @haruspex pointed out why you were getting an incorrect expression when the $x$ was negative. That's why it's really simpler to use a second expression to get the second point=there are otherwise too many minus signs floating around that should be with an absolute value sign.

Actually, I take that back. I just forgot to put the y in 2y on Wolfram

I ended up getting BOTH -6.20448 and .14629 as solutions. It seems that if I used THIS approach initially I would've got BOTH answers.

 

[haruspex]

Correct me if I am wrong, but I think there is only one solution above the apex of the triangle. The other solution is just below the base. You need to rewrite the equations for this second point.

How is that equation different?

 

[Charles Link]

How is that equation different?

See post 40. Ideally a single expression is all you need to get both solutions, but this one gets complicated by absolute value signs, etc. I think if you square both sides of the expression in post 40, you do get the second point $y=-.146$ as well.

 

[haruspex]

See post 40. Ideally a single expression is all you need to get both solutions, but this one gets complicated by absolute value signs, etc. I think if you square both sides of the expression in post 40, you do get the second point $y=-.146$ as well.

Ok, we were saying the same thing in two different ways. Where I expressed it as the equation being wrong, you expressed it as needing two versions of the equation.