Finding potential change with field E = 2xî - 3y²j N/C

[thepassenger48]

Homework Statement

An electric field is given by E = 2xî - 3y²j N/C. Find the change in potential from the position RA = î - 2j M to RB = 2î + j +3k m.

Homework Equations

∆V = VA - VB = - ∫ E • ds (definite integral from A to B)
Or in a uniform field, simply:
∆V = VA - VB = ±Ed

The Attempt at a Solution

I know from the solutions manual that the answer is +6 Volts.

I began by finding ∆R = RB - RA = i + 3j + 3k m.
That can easily be converted into ∆R = √19 right?

My problem comes in when I have to deal with E = 2xî - 3y²j N/C... what do I do with the square on the y?

Thank you once again.

 

[cepheid]

Have you had much experience with line integrals? Oh well...let me address specific aspects of your solution attempt first:

The Attempt at a Solution

I know from the solutions manual that the answer is +6 Volts.

I began by finding ∆R = RB - RA = i + 3j + 3k m.
That can easily be converted into ∆R = √19 right?

Not exactly. Be careful! $\mathbf{r_A}$ and $\mathbf{r_B}$ are vectors! So you were correct in stating that:

$$\mathbf{\Delta r } = \mathbf{r_B - r_A} = \mathbf{\hat{i}} + 3\mathbf{\hat{j}} + 3\mathbf{\hat{k}}$$.

But it doesn't make sense to say that $\mathbf{\Delta r } = \sqrt{19}$, because you're equating a vector to a scalar! Really, what you calculated was the magnitude of $\mathbf{\Delta r }$:

$$|\mathbf{\Delta r}| = \sqrt{19}$$

Unfortunately, I don't think that this is going to be of much use to you in solving the problem! I wouldn't start with calculating the difference between the two vectors, because to calculate the potential difference, you need to calculate the line integral of the electric field over some path between $\mathbf{r_A}$ and $\mathbf{r_B}$. You correctly wrote:

$$\Delta V = V(\mathbf{r_B}) - V(\mathbf{r_A}) = - \int_{\mathbf{r_A}}^{\mathbf{r_B}}{\mathbf{E} \cdot d\mathbf{l}}$$

But then...

My problem comes in when I have to deal with E = 2xî - 3y²j N/C... what do I do with the square on the y?

...this indicates you're stuck. But if you haven't calculated too many line integrals or taken vector calculus, don't despair. There is a half-assed, "physics" way to interpret it

Think of $d\mathbf{l}$ as an infinitesimal vector that points "along" the direction of the path at any point on the path. Then:

$$d\mathbf{l} = dx\mathbf{\hat{i}} + dy\mathbf{\hat{j}} + dz\mathbf{\hat{k}}$$

Therefore, you can calculate:

$$\mathbf{E} \cdot d\mathbf{l} = (2x\mathbf{\hat{i}} - 3y^2 \mathbf{\hat{j}}) \cdot (dx\mathbf{\hat{i}} + dy\mathbf{\hat{j}} + dz\mathbf{\hat{k}} ) = ?$$

your integral will turn into two integrals, one wrt x and the other wrt y, and you're off to the races.

 

[cepheid]

Uh, looking at the question again, I think I made the further assumption that since the E-field is a conservative vector field, then the answer is path independent, so you can simply use your starting and ending x & y points as limits of integration for each integral (i.e., take the straight line path).