Finding the Electric Potencial

1. Nov 8, 2009

kacete

1. The problem statement, all variables and given/known data
Given the Electric Field $$\vec{E}=40xy \vec{u_{x}} + 20x^{2} \vec{u_{y}} + 2 \vec{u_{z}} V/m$$ find the $$V_{PQ}$$, where $$P(1,-1,0)$$ and $$Q(2,1,3)$$.

2. Relevant equations
I know that:
$$V_{PQ}=-\int^{P}_{Q}\vec{E}d\vec{L}$$
I'm just confused, how to use it besides a straight line following a single direction $$\vec{u_{x}}$$ for example.

3. The attempt at a solution
I tried converting to spherical coordinates resulting in an equal confusion.
Solution (at the end of the book): $$106V$$.

Last edited: Nov 8, 2009
2. Nov 8, 2009

darkpsi

You could think of it as a vector connecting the two points P and Q. Thus the "straight line" would be equal to the distance between the points in the direction pointing from P to Q.

3. Nov 8, 2009

lubuntu

Notice you can drop the z component from the calculation because there is no change in it from Q to P.

4. Nov 8, 2009

kacete

I realized I made a typo, P is in fact (1,-1,0).

I thought of something like that, but if I integrate everything the result is a volume and not a line. In which direction should the integration be?

5. Nov 8, 2009

darkpsi

The direction would be composed of three different vectors. (Namely the vectors corresponding to the changes in the x, y, and z directions)

6. Nov 8, 2009

kacete

How about the beginning and end of the integral? Will it be $$0$$ and $$\sqrt{14}$$ (the module of the vector $$\vec{PQ}$$)?

7. Nov 8, 2009

darkpsi

Not quite. It would be easier if you changed the variables into just one variable and then you could integrate from, say, Px to Qx. I found the correct answer that way

8. Nov 8, 2009

kacete

How do you change the variables into just one?

9. Nov 8, 2009

darkpsi

Well you have to find how y and z vary with, say, x. If your electric field was just
E= xx+yy+zz then you could think of it as a triple integral and integrate over the three vectors that combine to create the vector u stated in your problem. But since your y-dependents vary with x, you have to find some way to relate the variables because your can't integrate a changing x as ∫xdy

10. Nov 8, 2009

kacete

Do you mean finding the vector component of the electric field in the direction of ux? For example.

11. Nov 8, 2009

darkpsi

Well it tells you the component in the x direction, it's 40xy. Maybe your confusing ux as having an actual magnitude? The magnitude is the 40xy. ux is simply saying the electric field (40xy) in the x direction, because the electric field is the same direction as vector u. To be accurate they really shouldn't even have a vector symbol above them but rather a hat because they are unit vectors. If you draw a simple grid and plot the two point (1,-1) and (2,1) you can see how y depends on x. Same goes for z. And then when the y's and the z's are in terms of x, you take the derivatives to get the dy and dz. Then you should have everything in terms of x's and dx's. Combine like terms and integrate.

12. Nov 8, 2009

kacete

I believe I understand now. Thank you very much for your help, I will try that.

13. Nov 8, 2009

gabbagabbahey

The electrostatic potential is path independent, and so the integration can be done over any path from $\textbf{P}$ to $\textbf{Q}$; not just the straight line path that you seem to imply is necessary.

Personally, I would break the path into three sections...one where only 'x' changes (so that $d\mathbf{\ell}=dx\mathbf{u}_x$), another where only 'y' changes (so that $d\mathbf{\ell}=dy\mathbf{u}_y$) and then a third where only 'z' changes (so that $d\mathbf{\ell}=dz\mathbf{u}_z$)

Of course, using the straight line from $\textbf{P}$ to $\textbf{Q}$ works as well, but it requires that you first parameterize that line.

14. Nov 8, 2009

kacete

Like this?

$$V_{PQ}=-\int^{P}_{Q}\vec{E}d\vec{L}=-(\int^{2}_{1}E\vec{u_{x}}dL\vec{u_{x}} + \int^{1}_{-1}E\vec{u_{y}}dL\vec{u_{y}} + \int^{3}_{0}E\vec{u_{y}}dL\vec{u_{y}})$$

15. Nov 8, 2009

gabbagabbahey

Your limits of integration are backwards,

$$V_{\textbf{PQ}}=-\int_{\textbf{P}}^{\textbf{Q}}\textbf{E}\cdot d\mathbf{\ell}=-\int_1^2 \textbf{E}\cdot dx\textbf{u}_x -\int_{-1}^1 \textbf{E}\cdot dy\textbf{u}_y-\int_0^3 \textbf{E}\cdot dz\textbf{u}_z$$

I suggest you draw a quick sketch of this path (remember you are starting at $\textbf{P}$ and finishing at $\textbf{Q}$)....what are the values of $y$ and $z$ along the first section of the path?...what are the values of $x$ and $z$ along the second section of the path?...what are the values of $x$ and $y[/itrex] along the final section of the path? 16. Nov 8, 2009 darkpsi Yes but then when you dot E with dL you'll get VPQ = ∫∫∫ 40xy dx + 20x2 dy + 2 dz Am I missing something or does changing of variables need to take place? 17. Nov 8, 2009 gabbagabbahey You are missing something....draw out the path i suggested....along the first section, [itex]y$ and $z$ will have a definite value...along the second section, $x$ and $z$ will have a definite value and along the last section, $x$ and $y$ will have a definite value.

Also, a path integral is quite different from a triple integral (if you are studying from Griffiths, I suggest you take a look at section 1.3.1!).

18. Nov 8, 2009

kacete

Right! I understand it now. Yes, it's easier than to parametrize the line. I didn't think of that, but it makes sense. Thank you! I will try it.

19. Nov 8, 2009

darkpsi

Well just for future reference, when couldn't I change everything to one variable to solve?

20. Nov 8, 2009

gabbagabbahey

Any curve can be written in terms of a single parameter (variable), so it is always possible to parametrize the straight line joining the two endpoints and do a single integration....However, it is often easier to choose a different path and avoid the parameterization process.