Finding the point where Coulomb forces sum to zero

[rickyw2777]

Homework Statement

Three identical point charges q are fixed at the vertices of an equilateral triangle with side length, a. Then, one of the charges is instantly changed to −q, and the system is released.
Task:
• Find analytically the equilibrium position of a positive test charge along the line of symmetry, where the net electric force on it is zero.
• Determine whether this equilibrium is stable or unstable.

Relevant Equations

F=kq1q2/r^2

I have tried to solve this problem by using the Columb's Law. Basically, I draw the graph and attempted to find the point where sum of the forces are zero. Here is what I have. Is there anything wrong in my working? Why can't I solve d in terms of a?

 

[BvU]

Interesting hieroglyphs! Could you provide a transcript ?

$\$

 

[haruspex]

The wording is very strange. What is the relevance of the charge being "instantly" negated, or of the system being "released". From what? From being fixed, so that it all flies apart?

Putting that aside, if your final equation is right then it gives a sixth degree polynomial which looks unlikely to have an algebraic solution. (Your sign handling looks questionable.)
You can check your algebra by considering the case where all three charges are q. You know what the solution to that is.
Or you can get a numerical solution and see whether it looks about right. Where should the equilibrium point be, roughly?

 

[BvU]

Interesting hieroglyphs! Could you provide a transcript ?

What I mean is a clear drawing and legible text. Like

Dug up an old egyptology book but couldn't find

. Physics book suggests $F_res=0$. Maybe, maybe not. y components cancel, and then

would be translated to $$F_{14,x}+F_{24,x}+F_{34,x}=0$$ in the drawing above. Give or take a minus sign....

Divide by $q_4$, let $kq_1=kq_2=1$ and the math problem looks like $$
{2\cos\theta\over r_{14}^2 }-{1\over r_{34}^2} =0, $$
which is quite comparable to

Except for the minus sign.... (from $q_3!$). Check it out: the $\cos$ terms should act in the opposite direction wrt the $r_{34}$ term !

Keeping things simple is for the faint-hearted, so you introduce $d=r_{34}$ and get

All positive !


Changing + to - gives an equation that agrees with a numerical result.

I can't find an analytical solution. Sure it exists ?

it all flies apart

Would it ? I think I agree, but can't prove ...

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[kuruman]

Would it ? I think I agree, but can't prove ...

It would fly apart because if you find a point where the electric field is zero, the equilibrium cannot be stable with respect to displacements in all three Cartesian directions. This is guaranteed by Laplace's equation in which the sum of the three partial second derivatives of the potential is zero. This means that the largest (in magnitude) of the three has sign which is opposite to the other two. To go around that here, we need to confine the motion in the $xy$-plane, say by imagining the charges sliding without friction along the $x$ and $y$ axes.

I would approach this problem by finding the electric potential energy $U(x,y)$ of the three-charge configuration excluding the irrelevant test charge. Then I would look for a minimum of this potential energy along the desired axis.

 

[PeroK]

I would approach this problem by finding the electric potential energy $U(x,y)$ of the three-charge configuration excluding the irrelevant test charge. Then I would look for a minimum of this potential energy along the desired axis.

Will that result in an analytic solution?

 

[kuruman]

Will that result in an analytic solution?

If you're asking, I must have goofed and I think I have. My method gives the equilibrium position about which the three-charge configuration oscillates, not the additional test charge which I declared irrelevant without thinking. Back to the drawing board.

 

[rickyw2777]

措辞很奇怪。“立即”取消指控和“释放系统”有什么关系？从哪里？从被固定住，然后一切都散架？

撇开这一点不谈，如果你的最终方程正确，那么它给出的是一个六次多项式，看起来不太可能有代数解。（你的符号处理看起来有问题。）
你可以通过考虑三个电荷都为q的情况来检查你的代数知识。你知道这个问题的答案是什么。
或者你可以求一个数值解，看看结果是否正确。平衡点大概应该在哪里？

Thank you. I have just asked the instructor and he said that that is all he needs. So thank you all who replied to this.

 

[kuruman]

Thank you. I have just asked the instructor and he said that that is all he needs. So thank you all who replied to this.

Can you please specify where in @haruspex's statement,

is what your instructor said that he needs and what that is?