- #1
bowlbase
- 146
- 2
I was told it might be better to post this here.
The trick to this problem is the E field is in cylindrical coordinates.
##E(\vec{r})=Cs^2\hat{s}##
##\int E \cdot dA##
I tried converting the E field into spherical coords and I can find the flux that way but it is a complicated answer. The problem suggests keeping the field in cylindrical and doing the integral of the circle in cylindrical instead of spherical. I'm sort of lost on how I would do that. Would I have the limits of s be 0→R and z -R→R and ##\phi## the same as hat it would normally be?
I doubt it is that simple but since I've never tried to use non-optimal coordinates for an object I'm not entirely sure how I would go about this.
Also, this is the work I've done:
For example
##E=Cs^2\hat{s}##
##s=rsin(\theta)## and ##\hat{s}=sin(\theta)\hat{r}+cos(\theta)\hat{\theta}##
so ##E=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})##
##\int E \cdot dA=E4\pi r^2=4\pi r^2(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})##
The next step it asks me to calculate the divergence of E and then graph it on the sz plane.
I can do this with the original equation but I now have answers in two different coordinate systems. Which I suppose sounds fair since they did gave me two also.
##∇\cdot E=\frac{1}{s}\frac{∂}{∂s}(sE_s)##
##=\frac{C}{s}(3s^2)=3Cs##
Finally, it asks that I now do the integral ##\int (∇\cdot E) dV## to show that the two methods are equivalent. At first glance I would say they are not. So I probably made a mistake somewhere.
Homework Statement
The trick to this problem is the E field is in cylindrical coordinates.
##E(\vec{r})=Cs^2\hat{s}##
Homework Equations
##\int E \cdot dA##
The Attempt at a Solution
I tried converting the E field into spherical coords and I can find the flux that way but it is a complicated answer. The problem suggests keeping the field in cylindrical and doing the integral of the circle in cylindrical instead of spherical. I'm sort of lost on how I would do that. Would I have the limits of s be 0→R and z -R→R and ##\phi## the same as hat it would normally be?
I doubt it is that simple but since I've never tried to use non-optimal coordinates for an object I'm not entirely sure how I would go about this.
Also, this is the work I've done:
For example
##E=Cs^2\hat{s}##
##s=rsin(\theta)## and ##\hat{s}=sin(\theta)\hat{r}+cos(\theta)\hat{\theta}##
so ##E=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})##
##\int E \cdot dA=E4\pi r^2=4\pi r^2(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})##
The next step it asks me to calculate the divergence of E and then graph it on the sz plane.
I can do this with the original equation but I now have answers in two different coordinate systems. Which I suppose sounds fair since they did gave me two also.
##∇\cdot E=\frac{1}{s}\frac{∂}{∂s}(sE_s)##
##=\frac{C}{s}(3s^2)=3Cs##
Finally, it asks that I now do the integral ##\int (∇\cdot E) dV## to show that the two methods are equivalent. At first glance I would say they are not. So I probably made a mistake somewhere.