Electric Potential question

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RESmonkey

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May 6, 2007
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I got the answer. It was easier then I was making it out to be. For some reason which I'll figure out tomorrow, I'm not supposed to care about direction here, at all. it's wierd because E field required directional computation.
 

eLiu

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Jun 4, 2001
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I thought we don't do hmwk questions on this board... and this is basic e&m to boot.

Anyway, I'll be nice. Do you know how to calculate the electric potential from 1 point charge? (A point in 3 dimensions that is.) You do, since it's on your formula sheet: V(r) = k*q/r in spherical coordinates. Where does this come from? The electric potential satisfies poisson's equation: laplacian(V) = rho/epsilon_0, where rho is the charge density. For the case of point charges, rho is linear combination of dirac-delta function. If you've had some ODEs, try to solve: 1/r^2 * diff(r^2 * diff(V(r),r),r) = delta(r)/epsilon_0 (the left-hand side is the nonzero part of the laplacian in spherical coords); the answer is the same V(r) i wrote earlier.

The thing to understand here is that Poisson's Eqn is linear. So solutions are additive. Thus if I have 3 point charges, I don't have to solve another differential equation to find the potential. I can just add up 3 independent solutions for 1 point charge! (linear superposition might ring a bell.) At some level, you knew this b/c you had to exploit linearity to solve the other parts of your problem.

The steps needed to find the electric potential at the origin should be obvious from here.

Also I should point out that this question is actually extremely vague. They don't specify the dimensionality of the problem. A point in 2D does not have the same potential as a point in 3D (a 2D "point" being equivalent to a line of charge in 3D). Basic e&m classes omitting this detail have always irked me.

To see this, find the potential for a point in 2D: solve 1/r*diff(r*diff(V(r),r),r) = delta(r)/epsilon_0 (left-hand side is the nonzero part of the laplacian in cylindrical or polar coords).

Edit: in response to your last comment... the potential is a scalar quantity. It is valid when curl(E) = 0, i.e. E is irrotational. Then you can write E = -gradient(V). And using Gauss's Law, we can write divergence(E) = -laplacian(V) = rho/epsilon_0.

So if you take my math on faith, then it should be obvious that E is a directional while V is scalar, since E = -gradient(V).

But why does that make sense? Well, the electric potential comes from: V = -integral(E dot dL), where E and dL are vector quantities. It's the energy needed to move charge through an E-field starting at point A, going to point B, and following some path. E is the electric field. dL is an incremental length. At the most general level, the integral there depends on the path taken through space.

But that should ring some alarm bells for you. I'm sure you learned in class that E-fields are conservative, path-independent, etc. What happened? Since E is irrotational (and the domain is "simply-connected"), the integral defining V is path-independent; E is conservative.

So if E were NOT conservative, then you could not define a single "global" potential! In fact the potential would then depend on the path you took. Luckily for us, E is conservative, and V is path-independent--no direction needed.

NOTE: E is only conservative under some pretty strict assumptions. But these are the assumptions you work with in intro e&m. Specifically, if you allow magnetic fields (i.e. move the charges "quickly"), then curl(E) != 0 and you need to appeal to something much more complex than Poisson's equation.
 

RESmonkey

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I'm about to go to bed, but I'm soooo going to do that tomorrow (the diffeq. method, even tho its beyond the scope of this class hehe ). Might be the first time playing with an ODE outside of math class. Just to avoid confusion, the notation you used ---> 1/r^2 * diff(r^2 * diff(V(r),r),r) , what does that look like with the proper symbols?

Thanks
 

eLiu

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Jun 4, 2001
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Originally posted by: RESmonkey
I'm about to go to bed, but I'm soooo going to do that tomorrow (the diffeq. method, even tho its beyond the scope of this class hehe ). Might be the first time playing with an ODE outside of math class. Just to avoid confusion, the notation you used ---> 1/r^2 * diff(r^2 * diff(V(r),r),r) , what does that look like with the proper symbols?

Thanks

Sure :) To be honest a lot of this is stuff I only realized several courses after e&m, because things were not explained to me in a very mathematical setting & I wasn't on top of my game enough to figure it out on my own (yet).

My notation is as follows... V=V(r,theta,phi), but due to symmetry I shortened it to V=V(r); diff(V(?),?) means the partial derivative of V with respect to ?. I can't really write out nice symbols on this board... I could type latex syntax but I think it'd be easier to point you to wiki: http://en.wikipedia.org/wiki/Laplace_operator

Scroll on down to the section "Coordinate expressions". This is where the "point in 2D" vs "point in 3D" distinction I was talking about becomes important. A point in 2D has cylindrical symmetry in 3D: observe that the 2D "polar coordinates" expression and the 3D "cylindrical coordinates" expression are the same (omitting the d/dz term).

So for the equation in question, we're solving laplacian_in_spherical_coordinates(V(r)) = 1/r^2 * d/dr(r^2 * dV/dr)) = delta(r)/epsilon_0. (d denotes partial deriv)

And it'd a good exercise to solve laplacian_in_cylindrical_coordinates(V(r)) = delta(r)/epsilon_0 and see that you get the potential for a line of charge.

Edit: don't forget to apply some logical boundary conditions
 
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