Thread: Line Integral and Electric Potential

1. Line Integral and Electric Potential

Hey I've been having a lot of trouble with a question involving the calculating the potential difference between two points.

So I have the following equations for Electric field:

$E_x = 6x^2y$
$E_y = 2x^3 + 2y$

Where x and y are coordinates that are given to me and I have to find the potential difference between (0, 0) and (x, y).

So I know that $V_{(x, y)} - V_{(0, 0)} = \int_{(x, y)}^{(0, 0)} E \cdot dl$

But I don't know how to evaluate that "line integral" basically. Also what is my dl???

2. Never mind I just got the answer. It's quite an interesting method mind you.

3. Originally Posted by TrevorP
Never mind I just got the answer. It's quite an interesting method mind you.
Why don't you fill us in? Perhaps your solution will help someone else.

-Dan

4. For sure I'll put it in tomorrow...It's 12:40 here right now and I've got class at 8:30 so I'll due it tomorrow afternoon.

5. Originally Posted by TrevorP
Never mind I just got the answer. It's quite an interesting method mind you.
In the meantime, for those too enthralled to wait, let me guess at the main results: ......

You realised that the electric field $\vec{E}$ was conservative (by checking whether $\nabla \times \vec{E} = 0$) and therefore:

1. Could be expressed in the form $\vec{E} = \nabla V$, and

2. Is independent of the path between the two points.

Then you got $V = 2x^3y + y^2$ and hence found that $
V_{(x, y)} - V_{(0, 0)} = \int_{(0, 0)}^{(x, y)} E \cdot dl = 2x^3 y + y^2$
.

The details will be of interest to others.

6. Ok here is the long and drawn out method (that actually just gives the same result as Mr. Fantastic got.):

$\vec{E} = (6x^2y)\hat{i} + (2x^3 + 2y)\hat{j}$

$\vec{r} = tx\hat{i} + ty\hat{j} \ \ 0 \leq t \leq 1$

$\vec{E}(\vec{r}) = (6(tx)^2(ty))\hat{i} + (2(tx)^3 + 2(ty))\hat{j}$

$\tfrac{d}{dt} \vec{r} = x\hat{i} + y\hat{j}$

$\vec{E}(\vec{r}) \cdot \tfrac{d}{dt} \vec{r} = (6(tx)^2(ty))x + (2(tx)^3 + 2(ty))y = 6x^3yt^3 + 2x^3t^3y + 2y^2t$

$\vec{E}(\vec{r}) \cdot \vec{r} = (6(tx)^2(ty))tx + (2(tx)^3 + 2(ty))ty$

$V = \int_0^1 (6x^3yt^3 + 2x^3t^3y + 2y^2t) dt = 2x^3y + y^2$

It's a long and drawn out method that but it can be adapted to include the z axis as well as from different points. The challenge is just getting the right position vector.

7. Originally Posted by mr fantastic
In the meantime, for those too enthralled to wait, let me guess at the main results: ......

You realised that the electric field $\vec{E}$ was conservative (by checking whether $\nabla \times \vec{E} = 0$) and therefore:

1. Could be expressed in the form $\vec{E} = \nabla V$, and

2. Is independent of the path between the two points.

Then you got $V = 2x^3y + y^2$ and hence found that $
V_{(x, y)} - V_{(0, 0)} = \int_{(0, 0)}^{(x, y)} E \cdot dl = 2x^3 y + y^2$
.

The details will be of interest to others.
Let $E = \nabla V = \frac{\partial V}{\partial x} i + \frac{\partial V}{\partial x} j$.

Then:

$\frac{\partial V}{\partial x} = 6x^2 y$ .... (1)

$\frac{\partial V}{\partial Y} = 2x^3 + 2y$ .... (1)

Equations (1) and (2) are to solved simultaneously:

(1) => $V = 2 x^3 y + C(y)$ where C(y) is an arbitrary function of y.

Sub this result into (2):

$2 x^3 + \frac{dC}{dy} = 2x^3 + 2y \Rightarrow \frac{dC}{dy} = 2y \Rightarrow C = y^2 + K$

Therefore $V = 2 x^3 y + y^2 + K$.

You can always do this if E is conervative ......