# Curl and Divergence

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.

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
galipop said:
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?

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
Homework Helper
Gold Member
Dearly Missed
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 $$\Phi$$ ?

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).

Last edited:
HallsofIvy
Homework Helper
arildno said:
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.

Last edited by a moderator:
arildno
Homework Helper
Gold Member
Dearly Missed
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)

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.

Thanks guys....

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