2 identical charges @ x=a & x=-awhere does E field vanish?

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Homework Help Overview

The discussion revolves around determining the point or points along the x-axis where the electric field from two identical charges located at x=a and x=-a vanishes. Participants are exploring the conceptual understanding of electric fields and their interactions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the conditions under which the electric field cancels out at the origin due to the symmetry of the charges. Other participants question the necessity of algebraic verification versus conceptual reasoning.

Discussion Status

Participants are actively engaging with the problem, with some providing references to previous discussions and others offering insights into the reasoning process. There is a mix of conceptual exploration and attempts to clarify the mathematical approach needed for subsequent parts of the problem.

Contextual Notes

Some participants express confusion regarding the need for specific values in calculations and the implications of symmetry in the electric field analysis. There is also mention of the need to derive expressions for the electric field components and maximize them, indicating a focus on calculus without explicit values.

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Homework Statement


Determine at what point or points (not including points at infinity) the electric field vanishes.


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The Attempt at a Solution



Well, this seems to be a conceptual question and as I must be missing the concept, I don't know how to attempt to even think about it. Any help would be greatly appreciated.
 
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Ah, I didn't catch that thread - thank you!
However, quick question then - if my question says that my charges are identical and both the same distance away from the origin on the x-axis (a and -a), then because the charges are equal, wouldn't their fields be equal at the origin and thus cancel out, and so then the answer would be that the electric field vanishes at the origin because that's where the electric field of the charges is equal?

Please let me know if that is okay reasoning, or if it really is required to go through the algebra to deduce it. Thanks :)
 
That's how I would do it :D
 
Okay, cool. Thanks for directing me to that post - otherwise, I wouldn't have read the part that said 'E1 is equal to E2' and it wouldn't have clicked what my conclusion to my own question is ten.

If I could just get help with another part of this question..?
The next part asks to find an expression for the y component of the electric field on the y-axis as a function of y. So I just drew a point charge on the y-axis some distance up and made a triangle between the two Q charges and the point charge. Taking just the right triangle, I use sine to get sin([itex]\theta[/itex])=Ey/E. Then, using F = kQ2/r2 and r2 = sqrt(a2+y2)... I get
Ey = (2kQ2sin([itex]\theta[/itex])) / (q0 sqrt(a2+y2))

Is that the correct expression? I'm confused because then the next part asks to "Find the point(s) on the y-axis where the magnitude of Ey is a maximum." This means I need to find where the derivative is zero, so dEy = 0.
However, I don't know have any values for quantities. But I suppose all the variables are constant except for y, right? So now I just need to remember how to do calculus from back in the day.


My question here is just - is that the right way to go about it?
 
Your answer is a bit confusing, you should try to be more schematic... I mean, I think you got it almost right, but it lacks something, and thinking more schematically can and will help you.
This is what I mean:

1. You are looking at the electric field, so (though useful) you don't actually need the point charge q0.

Now, as you did implicitly, let's make some symmetrical analysis:
2. The situation is axial-symmetrical: so, we can calculate Ey of one charge and then (in this case) double it to consider the effect of the other one.

3. Now your turn: since you need the dependence of the electric field from the y-component, let's calculate Ey(y). Which is the basic equation? How do you get the y-component?

4. How do you calculate Ey depending only from y? (Here is the problem with your answer: your Ey depends on y, but also on teta! This makes taking the derivative a very complex task.)

5. You need to find the maximum... all right then, back to calculus! :D As you said, you need to find the Y so that dEy(Y)=0.

Hope it helps. You were almost there, so if you like you can just use your result (Ey = (2kQ2sin(θ)) / (q0 sqrt(a2+y2))) and consider only points 4 and 5 to go on.

Remember to replace sin(teta) with some y-dependent expression before deriving. This is the main thing to do.
 

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