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mitleid
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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?
Coulomb's Law
E = ke*q/r[tex]^{2}[/tex]
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[tex]^{2}[/tex])
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[tex]^{2}[/tex] = y[tex]^{2}[/tex] + a[tex]^{2}[/tex]. Therefor E(-) = Ke(-2q/y[tex]^{2}[/tex]+a[tex]^{2}[/tex]).
Now I have to break this down to find the y-component for E(-) which (I think) is just :
Ey(-) = sin[tex]\alpha[/tex]*E(-)
Since the ultimate goal here is to define two functions, should I define sin[tex]\alpha[/tex] in terms of y for integration purposes? Could I say sin[tex]\alpha[/tex]= y/r = y/(y[tex]^{2}[/tex]+a[tex]^{2}[/tex])[tex]^{1/2}[/tex]? 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?
Coulomb's Law
E = ke*q/r[tex]^{2}[/tex]
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[tex]^{2}[/tex])
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[tex]^{2}[/tex] = y[tex]^{2}[/tex] + a[tex]^{2}[/tex]. Therefor E(-) = Ke(-2q/y[tex]^{2}[/tex]+a[tex]^{2}[/tex]).
Now I have to break this down to find the y-component for E(-) which (I think) is just :
Ey(-) = sin[tex]\alpha[/tex]*E(-)
Since the ultimate goal here is to define two functions, should I define sin[tex]\alpha[/tex] in terms of y for integration purposes? Could I say sin[tex]\alpha[/tex]= y/r = y/(y[tex]^{2}[/tex]+a[tex]^{2}[/tex])[tex]^{1/2}[/tex]? 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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