Solving Constants a & b for Electric Field E - Curl & Divergence

[galipop]

Hi All,

Given electric field E=c(2bxy,x^2+ay^2), I need to determine the constants a and b such that CURL E = 0 and DIV E = 0. I'm also given a path from (0,0) , (1,0) and (1,1).

Ok so the curl = 0+0+cx(2-b) = 0
and the divergence = 2cy(b+a) = 0

How do I solve for a and b at this stage given so many variables?

Also how do I compute the potential in this case.

 

[Doc Al]

Ok so the curl = 0+0+cx(2-b) = 0
and the divergence = 2cy(b+a) = 0

How do I solve for a and b at this stage given so many variables?

First, recheck your expression for the curl.

Then ask yourself: what must "b" be to make the curl always zero (that is, for any value of x and y)? What must be the relation between "b" and "a" to make the divergence always zero?

 

[galipop]

ok so I redid the curl and got 2xc(1-b) = 0

therefore b=1 to satisfy that the curl = 0
and a=-1 so that the div = 0 :)

ok that was easy after all...


now how do I compute the potential?

 

[arildno]

Given potential $$\Phi$$ you must have:
$$\frac{\partial\Phi}{\partial{x}}=2bcxy$$
and
$$\frac{\partial\Phi}{\partial{y}}=c(x^{2}+ay^{2})$$
How can you use these equations to determine $$\Phi$$ ?

 

[arunma]

Remember also that when you integrate with respect to one variable, the "constant" will actually be a function of the other variable.

Actually, I tried working this problem out for you. But the answer I kept getting was slightly off from what the real potential should be (sorry, it's been a couple years since I took multivariable). In any case, the answer should be f(x,y) = (1/3)acy³ + cx²y. But definitely do arildno's method and make sure that you get this answer (professors don't react to well to answers without work, as I'm sure we've all experienced).

 

[HallsofIvy]

Given potential $$\Phi$$ you must have:
$$\frac{\partial\Phi}{\partial{x}}=2bcxy$$
and
$$\frac{\partial\Phi}{\partial{y}}=c(x^{2}+ay^{2})$$
How can you use these equations to determine $$\Phi$$ ?

Notice that by differentiating the first equation with respect to y, $\frac{\partial^2\Phi}{\partial x\partial y}= 2bcx$ and by differentiating the second equation with respect to x, $\frac{\partial^2\Phi}{\partial y\partial x}= 2cx$. Since, for any function with continuous second partials, the two mixed partial derivatives must be equal this is possible only if b= 1. Otherwise there is no $\Phi(x,y)$ that has those derivatives.

Integrating the first equation with respect to x, treating y as a constant, $\Phi= bcx^2y+ f(y)$ (the "constant" of integration may depend on y).

Differentiating THAT equation with respect to y, we get $\frac{\partial\Phi}{\partial y}= bcx^2+ \frac{df}{dy}$.

Comparing that with the second given equation, we must have
$bcx^2+ \frac{df}{dy}= cx^{2}+acy^{2}$.

In order for that to be true, the terms involving x must cancel which happens only for b= 1. Taking b= 1 we can cancel that term and have
$\frac{df}{dy}= acy^2$. Integrating that, $f(y)= \frac{1}{3}acy^3+ C$. Because f depends only on y, C really is a constant.
Putting those together, we have:
$\Phi(x,y)= cx^2y+\frac{1}{3}acy^3+ C$
as long as b= 1. If b is not 1, there is no such function.

 

[arildno]

HallsofIvy:
Since the first part of the exercise was to determine those values of a and b which gives zero curl(&divergence,) those are the ones that were implied to be used in my example.
(The existence of the potential has therefore been assured by these choices)

 

[arunma]

That's why I kept messing up when I tried to actually calculate the potential! I forgot that b must equal 1.

Also I completely forgot about Clairaut's theorem. Amazing how much stuff there is to forget in math.

 

[galipop]

Thanks guys...

Arunma, how did you know the solution without being able to derive it correctly?