# Electric Potential

## 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.

Related Introductory Physics Homework Help News on Phys.org
cepheid
Staff Emeritus
Gold Member
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 :rofl:

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
Staff Emeritus