Solving Electric Potential Difference in a Linearly Varying Field

[physics girl phd]

Homework Statement

I'm trying to do a problem two ways, and things aren't consistent, finding the electric potential difference in a linearly varying field.

The electric potential difference between two points is often summarized in texts as ΔV = Vf - Vi = - ∫ E⋅ds where the lower bound of integration is the initial point and the upper bound of integration is the final. It's my understanding the vector ds points from the initial point to the final point.

My example is a field with linear variation, pointing in the +i direction, with value E = -Bx , and the points we are concerned with are x1 and x2, which are both in the +x region. I would expect to be able to independently find V2-V1, and V1-V2, and find they are equal and opposite. But for some reason I'm not getting that. When the bounds of integration flip, and the path vector ds from one point to another flips, you get two negatives that cancel the effects, so you get the same ΔV, not an equal but opposite ΔV.

I attach my work.

From what I can tell looking at various texts I have (from introductory to intermediate... I haven't yet pulled out grad-level), their definitions of ds are often vague, and they don't change its direction in examples... can someone clarify this?

Homework Equations

ΔV = Vf - Vi = - ∫ E⋅ds where the lower bound of integration is the initial point (or reference) and the upper bound of integration is the final. It's my understanding the vector ds points from the initial point to the final point (but I could be wrong there).

The Attempt at a Solution

I attach my work. I get the correct delta V one way... but don't get the equal and opposite the other (which is from the further point inward.

 

[kuruman]

That's an often seen error that beginner's make with line integrals. The integrand here is $\vec E \cdot d\vec s$ where $\vec E = -Bx~\hat i$ and $d\vec s=dx~\hat i$. Then $\vec E \cdot d\vec s=-Bxdx$. The reversed limits of integration flip the sign. For future reference: when doing line integrals always use $d\vec s=dx~\hat i +dy~\hat j +dz~\hat k$ regardless of the integration limits, then form the dot product and then integrate.

On edit: Here is another way to look at it. Element $dx$ is by definition positive in the direction of increasing $x$. When you reverse the sense of integration, you cannot say that $dx$ is positive in the opposite direction because that's changing coordinate axes midway through the calculation. Once you pick a coordinate system you must stick with it. No wonder you get inconsistent results!

 

[physics girl phd]

I figured it had something to do with that, but all the standard derivations of this equation are through work done with a path in the same direction as the coordinate system, which simplified things in one way, but confused things in the other (and with work I always check energy and signs for correctness anyways, since it's so easy to flip things depending on if you think about what might be doing the work).

So would you think it would be best to define ds as the "standard" infinitesimal displacement vector that comes from the coordinate system, and the bounds as setting the path taken (which of course only depends on the end points in this conservative case)? Or how do you think to best say this?

 

[kuruman]

View the line integral as what it is, a sum of a whole bunch of infinitesimally-small scalar dot products. Because it is a sum of scalars, the line integral is independent of your choice of coordinate system. You choose a coordinate system only because this choice facilitates writing the dot product. That is why I strongly recommend that one write the vector field $\vec F$ and the line element $d\vec s$ first in unit vector notation, then take the dot product formally and finally add the scalars. Do that and you can't go wrong. You must think of $d\vec s$ as an element of a "position" vector, not as an element of a "velocity" vector that has to point in the direction of motion from the starting point to the ending point.

In your example, you were given a one-dimensional field $E=-Bx$. Assuming that $B$ is positive, it is implicit that this field points in the negative x-direction, i.e. opposite to the direction of increasing $x$ which also happens to be the direction of $d\vec x$ should you choose to represent it as a vector and draw it as an arrow. Thus, the two vectors are at 180^o, $\vec E \cdot d\vec s=-E~dx=-Bx^2dx$. Doing the integral means adding a whole lot of negative numbers if you are going in the direction of increasing $x$ and a whole lot of positive numbers if you are going in a direction of decreasing $x$. The limits of integration take care of that automatically because they determine which way to take the difference between the end points of $F(x)=-\frac{1}{3}Bx^3=-V(x)$.

 

[physics girl phd]

So perhaps best to jointly state that in the original text-equation "-E ⋅ ds" is a little change in potential along a small position change defined by the coordinate system?

The limits ultimately define where that position change is in the coordinate system, and ds defines the step size in position changes along the path... We could take discrete little steps and have in-between limits to have numbers that we add to get our overall change, but we can in many cases make ds infinitesimal and smoothed out in our eventual integration calculation with only final limits (unless E suddenly changes somewhere say). Of course we can take various paths because E is a conservative field (which means we can do this integral to find the overall delta V), but ultimately we'll want to probably chose a friendly coordinate system/origin for our mathematics... and make sure that E and s (position) are defined from that, and ds is defined from s.

I think I feel comfortable with that (unless you see weirdness or somewhere that lacks clarity). If E is aligned with ds and the bounds of integration we drop in potential (confirming E is a "downward slope" of V; a free-to-move test "+" charge placed in the field loses potential energy as it is naturally driven by the field through a small change in position). If instead we chose an initial point as a reference, say far away, or at another radius (as in spherical and cylindrical systems), our path (from position changes that add to final bounds of integration) is often against E (if it's radially out from a positive charge or positively charged wire), and we have an increase in V, but that increase (or + sign) naturally comes out after the integration as we set it up and is what we will look for.

My calc 3 class was just pretty weakly absorbed (a long time ago), and I really try to make sure every term in an equation gets clearly defined because my calc 3 is really learned through physics. At my stage now, it's not just getting the right answer (if that was the case I'd slap on absolute values and throw in a sign at the end via a physical argument), but really about understanding how to make the whole process clear. I've become insistent on precisely defining things from the get-go, and I wasn't happy with how ds not really clearly defined. Then things weren't matching up with how I thought it was defined.

(I'm just noticing in my above example I have my expectation written wrong, the top is "+" as expected and the bottom is "not -," but I'd gone through a lot of variations of various E's in the past day and am rather brain-fried). Relating the path o both ds position changes and intermediate bounds makes much more sense.

 

[kuruman]

So perhaps best to jointly state that in the original text-equation "-E ⋅ ds" is a little change in potential along a small position change defined by the coordinate system?

Not quite. $-\vec E \cdot d\vec s$ is the potential difference between the two ends of a small element of length $ds$. Now a difference, as you know, is always what comes later minus what comes first, the value at the ending point minus the value at the starting point. The direction of integration along the path determines which of the two points comes first and which comes later.