Electric Potential

  • #1

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.
 

Answers and Replies

  • #2
cepheid
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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! [itex] \mathbf{r_A} [/itex] and [itex] \mathbf{r_B} [/itex] are vectors! So you were correct in stating that:

[tex] \mathbf{\Delta r } = \mathbf{r_B - r_A} = \mathbf{\hat{i}} + 3\mathbf{\hat{j}} + 3\mathbf{\hat{k}} [/tex].

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

[tex] |\mathbf{\Delta r}| = \sqrt{19} [/tex]

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 [itex] \mathbf{r_A} [/itex] and [itex] \mathbf{r_B} [/itex]. You correctly wrote:

[tex] \Delta V = V(\mathbf{r_B}) - V(\mathbf{r_A}) = - \int_{\mathbf{r_A}}^{\mathbf{r_B}}{\mathbf{E} \cdot d\mathbf{l}} [/tex]

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 :rofl:

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

[tex] d\mathbf{l} = dx\mathbf{\hat{i}} + dy\mathbf{\hat{j}} + dz\mathbf{\hat{k}} [/tex]

Therefore, you can calculate:

[tex] \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}} ) = ? [/tex]

your integral will turn into two integrals, one wrt x and the other wrt y, and you're off to the races. :smile:
 
  • #3
cepheid
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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).
 

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