Field and Potential of 3 charges conceptual problem

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The discussion focuses on calculating the electric field (E) and electric potential (P) at points along the positive y-axis due to a configuration of one positive charge (+2q) and two negative charges (-q). The x-component of the electric field is zero, while the y-component is derived from the contributions of the positive charge and the negative charges, requiring trigonometric considerations for accurate calculations. The participants explore the relationship between the electric field and potential, specifically whether the y-component of the field satisfies the equation Ey = -dP/dy. They emphasize the importance of integrating the equations for Ey and P, noting that potential is simpler to work with as a scalar. The discussion highlights the challenges of applying calculus concepts in deriving these functions.
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One positive (+2q) and two negative (-q) charges are arranged as displayed in figure 1. Calculate electric field E and electric potential P at points along the poitive y-axis as functions of their coordinate y. What is the direction of E at those points? In your results, does y-component Ey of the field satisfy Ey = -dP/dy? Is it supposed to satisfy?

figure1.jpg


Coulomb's Law
E = ke*q/r^{2}

First of all, the x-component of the field at any point along the y-axis is zero, since -q(-) and -q(+) cancel out one another, and 2q provides no field in the x-direction.

I know the field at any point will be equal to the field from 2q (positive y) minus the two y-components from the other two particles (negative y). The contribution from the positive particle is simple.

Ey(+) = ke(2q/y^{2})

The other contributions require a little trigonometry, which I'm hoping I've done correctly..

Assuming r is equal to the distance from -q to y (hypotenuse), r^{2} = y^{2} + a^{2}. Therefor E(-) = Ke(-2q/y^{2}+a^{2}).

Now I have to break this down to find the y-component for E(-) which (I think) is just :

Ey(-) = sin\alpha*E(-)

Since the ultimate goal here is to define two functions, should I define sin\alpha in terms of y for integration purposes? Could I say sin\alpha= y/r = y/(y^{2}+a^{2})^{1/2}? Oof, things are getting rusty...

My gut says I will have to integrate the equations for Ey(+) and Ey(-), and the difference between them will be my function for the field. I haven't really started at the potential equation yet... figured I would check to see if I'm headed the right direction first. Any advice?
 
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It's much easier the other way around. Obtain an expression for the potential, which is simpler to work with being a scalar.

You can then obtain the electric field with a gradient operator (in other words Ex=-dV/dx Ey=-dV/dy, etc. (partial derivatives)).
 
Potential = Ke (q/r), but r = \sqrt{y^{2}+a^{2} for the two negative charges.

So will the total P be Ke(2q/y) + Ke(-2q/\sqrt{y^{2}+a^{2})?

When I derive this I get something like Ke(dq/y^{2}) -Ke(dq/2(y^{2}+a^{2})^{3/2})

I see! This is the same as Ey with substituted sin like I was asking. The only issue I see is when I derived I got a 2 in the denominator...
 
Don't forget the chain rule... :wink:
 
hahah... I can't help it, my mind refuses to retain calculus methods.
 
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