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Revision as of 12:18, 8 November 2017
Edited by Josh Whitley Spring 2016, Emilie Pourchet Fall 2016, William Nute Fall 2017
CLAIMED BY WILLIAM NUTE FALL 2017
Potential difference is a scalar quantity; this means it has no direction and only a magnitude. Moreover, potential difference is also a "path independent" quantity. This means that given two different observation locations, the potential difference between those two locations will always be the same no matter what path is taken between them (similar to work, only here we use electric fields rather than forces). The path independent nature of potential helps to simplify electric potential calculations and enables us to solve problems more easily by allowing us to define paths to simplify calculations.
Why is Potential Path Independent?
Before discussing the path independence of electric potential, it can help to review the concept and rationalize why it might be a path independent quantity based on its characteristics.
Electric potential is essentially a value created to physicists to aid in their calculations. In a similar vein to electric field, which is a convenient way to generalize the force exerted on a particle of unit charge place at a given location, potential is a way to generalize the electric potential energy of a particle of unit charge placed at a given location. By nature, electric potential requires 2 points to define it (hence the term "potential difference"), so you will often hear something like "potential difference between A and B". If the potential of a single point is ever mentioned, it's typically in reference to the potential difference between that point and an arbitrary point infinitely far away from all sources of electric field.
So, in essence, electric field, electric potential, and potential difference are the "electric analogues" of force, potential energy, and work. Think for a minute about the relationship between force and work. Work is a path-independent quantity; no matter which way you push an object, how circuitous or direct your route is, the work required to get from one point to another is always the same. Given the analogy, it's not much of a stretch to conclude that potential difference, the electric analogue of work, is similarly a path independent quantity.
Another way of seeing the path independent nature of V is to look at the equation itself! The definition of potential is formally written as
When you expand out the dot product, you get an expression
which breaks up the potential difference into the potential difference across the x component, then the y component, then the z component. If you can break it into 3 different paths this way, then why not any other way?
Path Independence
Now that we know some logical reasoning behind why potential has the quality of path independence, what is it?
Path independence is a very useful quality for a value to have. If a value is path independent, then by definition we don't care about any of the information along a path that a particle takes; in other words, we only care about the endpoints of that path. This is a property that allows us to define [math]\displaystyle{ V_{AB} = V_B - V_A }[/math] in the first place.
Moreover, path independence allows us to do some tricks when calculating potential difference. Calculating potential difference for a path through constant electric field is much simpler than calculating potential difference for a path through changing electric field - the latter requires you to integrate an electric field expression, which often times can get messy. By taking advantage of path independence, it's often possible to draw a path from one point to another such that there's constant electric field, and in cases where you do have a changing field, drawing an effective path is often a critical part of getting the solution.
In more formal notation, you could see something like this:
where 1, 2,..., n are all different points.
Visually, this would look like splitting up a path going straight from A to B into a path that goes from A to n, then to n-1, so on and so forth until point 2, then 1, then finally B.
A Computational Model
VPython Model for Path Independence
The above is a visual and computational representation of the idea of path independence. Two different particles, one blue and one red, will begin in the bottom corner and take different paths to reach the top corner. The blue particle will take a direct straight-line path while the red particle will take a more indirect path. After each step forward (done by clicking on the scene), the change in potential from the last step and total change in potential will be printed for each particle. Note that at the end of the program, when both particles have reached the top corner, their total changes will be the same! Point A is marked by a green sphere, Point B is marked by a red sphere, moving negative particle 1 is marked by a blue sphere and trail, while moving negative particle 2 is marked by a white sphere and trail. Remember to scroll down in the display window to make sure you've seen the most recent printed value!
Examples
Example 1
This a basic example of using path independence to simplify the calculation of potential difference.
Here we have 2 locations, A and B, that are on the two plates of a capacitor, and we're trying to find the potential difference [math]\displaystyle{ \Delta V = V_B - V_A }[/math]. We're given the field in between the plates is [math]\displaystyle{ \vec{E} = \langle E,0,0\rangle }[/math]. The diagram on the right shows a way that we can break up this path into the [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] directions. This is an example of splitting up the path into the dot product, as discussed above.
Using path independence, we can get the equation [math]\displaystyle{ V_B - V_A = (V_B - V_C) + (V_C - V_A) }[/math]. These two smaller potential difference will be easier to handle, since path is always on axis with field.
From A to C:
[math]\displaystyle{ \Delta V = -E_xl }[/math]
[math]\displaystyle{ \Delta V = -Es }[/math]
From C to B:
[math]\displaystyle{ \Delta V = -E_yl }[/math]
[math]\displaystyle{ \Delta V = 0 }[/math]
To calculate total potential difference, we'll just add our two values together.
[math]\displaystyle{ \Delta V=V_B - V_A = (V_B - V_C) + (V_C - V_A) }[/math]
[math]\displaystyle{ \Delta V = -Es + 0 }[/math]
[math]\displaystyle{ \Delta V = -Es }[/math]
While in this example, the dot product calculation of [math]\displaystyle{ \Delta V = -\langle E_x\Delta x, E_y\Delta y\rangle }[/math] could've been applied pretty easily without breaking the path up, you'll see in the next example that defining a convenient path will often allow you to solve otherwise unsolvable problems.
Example 2
We're given the above diagram of a dipole and 2 points A and B, the former on axis and the latter on the perpendicular axis, each a distance [math]\displaystyle{ d }[/math] away from the center of the dipole. Each charge in the dipole has magnitude [math]\displaystyle{ q }[/math] with separation distance [math]\displaystyle{ s }[/math] between them. For a dipole, we've only learned equations for a field on axis and perpendicular to the axis, so at first glance it may seem impossible to find the potential difference [math]\displaystyle{ \Delta V = V_B - V_A }[/math]; after all, the field of a dipole along that path is quite irregular. However, there is a way to solve the problem using path independence.
You might be tempted to try to draw a path from A to the center of the dipole, then to B, but this will create some problems. One, you won't be able to use the approximate dipole formulas when extremely close to the dipole; and two, you'd end up with a division by zero when the radius goes to zero. Instead, we can cheat a bit by traveling around from on axis to perpendicular axis, while avoiding the complicated fields by making this move at a distance infinitely far away. Since the field of a dipole at all infinitely distant locations is zero, we can travel between "different" infinities (like one on axis and one on the perpendicular axis) without any potential difference between the two locations.
Our equation for this path is going to look something like the following:
Since the field for a dipole varies with distance, we're going to have to integrate the field formula.
However, we can see that the dot product in the first integral evaluates to zero; that is, the field on the perpendicular axis of the dipole is perpendicular to the path we're taking, so the potential difference for that integral is zero. Now, we can just solve the second integral, plugging in the on axis approximation for a dipole.
[math]\displaystyle{ \Delta V = -\int_{A}^{\infty}\langle \frac{1}{4\pi\epsilon_0} \frac{2qs}{r^3},0,0\rangle\cdot\langle dx,0,0\rangle }[/math]
[math]\displaystyle{ \Delta V = -\frac{qs}{2\pi\epsilon_0}\int_{d}^{\infty} \frac{1}{x^3}dx }[/math]
[math]\displaystyle{ \Delta V = \frac{qs}{2\pi\epsilon_0}\bigg(\frac{1}{2x^2}\bigg)\bigg|_{d}^{\infty} }[/math]
[math]\displaystyle{ \Delta V = \frac{qs}{2\pi\epsilon_0}(0) - \frac{qs}{2\pi\epsilon_0}\bigg(\frac{1}{2d^2}\bigg) }[/math]
[math]\displaystyle{ \Delta V = -\frac{qs}{4\pi\epsilon_0d^2} }[/math]Relation to Multivariable Calculus
Students who have taken or are taking multivariable calculus may have studied path independence in a different context before. Path independence is heavily related to the Fundamental Theorem of Line Integrals, potential functions, and conservative vector fields.
The Fundamental Theorem of Line Integrals states that if a vector field is conservative (that is, if a potential function exists for that field), then the line integral of a path through that field is equivalent to the difference in the potential function evaluated at the endpoints of that path.
In physics terms, the field that we're integrating is electric field. Electric field will always be a conservative vector field since, by definition, it has no means of dissipating energy (when all that's present is an electric field, energy will be conserved). This means that field has a potential function, which is the aptly named value electric potential (well, more precisely, a negative one multiplied by electric potential). So, rewriting the FTLI with physics values, we have:
Hopefully the above should look familiar - it's how we've been defining potential difference ever since we started learning it!
A consequence of this property is that a line integral in a conservative field along a closed path (i.e. the start and end points are the same) will be 0. This parallels the idea that round-trip potential difference equals 0, which is incidentally the same as Ampere's Law for Electric Fields.
See also
https://en.wikipedia.org/wiki/Voltage#See_also
https://en.wikipedia.org/wiki/Electric_potential
https://www.insidescience.org/content/soccers-electric-potential/1022
http://jes.ecsdl.org/content/147/11/4263.abstract
http://farside.ph.utexas.edu/teaching/302l/lectures/node32.html
http://faculty.cua.edu/sober/611/PATHIND.pdf
http://scienceline.ucsb.edu/getkey.php?key=4026
References
Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.
Chapter 16 Webassign Review