Path Independence of Electric Potential: Difference between revisions

Latest revision as of 16:14, 16 April 2023

Edited by Josh Whitley Spring 2016, Emilie Pourchet Fall 2016, William Nute Fall 2017

CLAIMED BY WILLIAM NUTE FALL 2017

See the picture drawn below. It is a clear example of path independence. If someone were to travel from point A to C to D to B, it would essentially be the same as traveling straight from A to B when we think about potential.

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

[math]\displaystyle{ \Delta V = -\int\vec{E}\cdot d\vec{l} = -\vec{E}\cdot \Delta \vec{l} }[/math] for constant [math]\displaystyle{ \vec{E} }[/math]

When you expand out the dot product, you get an expression

[math]\displaystyle{ \Delta V = -\langle E_x\Delta x, E_y\Delta y, E_z\Delta z\rangle }[/math]

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.

Mathematical Model

In more formal notation, you could see something like this:

[math]\displaystyle{ V_{AB} = (V_B - V_1) + (V_1 - V_2) + ... + (V_{n-1} - V_n) + (V_n - V_A) }[/math]

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.

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.

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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

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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.

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Our equation for this path is going to look something like the following:

[math]\displaystyle{ \Delta V = V_B - V_A = (V_B - V_\infty) + (V_\infty - V_A) }[/math]

Since the field for a dipole varies with distance, we're going to have to integrate the field formula.

[math]\displaystyle{ \Delta V = -\int_{\infty}^{B}\vec{E}\cdot d\vec{l} - \int_{A}^{\infty}\vec{E}\cdot d\vec{l} }[/math]

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}\vec{E}\cdot d\vec{l} }[/math]

[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.

[math]\displaystyle{ \int_{a}^{b}\vec{f}\cdot d\vec{r} = F(\vec{r}(b)) - F(\vec{r}(a)) }[/math] where [math]\displaystyle{ \nabla F = f }[/math]

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:

[math]\displaystyle{ -\int_{a}^{b}\vec{E}\cdot d\vec{l} = V(B) - V(A) }[/math]

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.

Connectedness

1. Path independence is a topic that originally stems from multivariable calculus, which has always been one of my favorite subjects in mathematics. I loved studying the connection of so many different topics to culminate in concepts like Green's Theorem, Stoke's Theorem, and as mentioned in this page, the Fundamental Theorem of Line Integrals. Multivariable calculus has a huge connection to electromagnetism, and especially Maxwell's Equations.

2.Thermodynamics and electric potential are both topics that rely greatly on path independence and are intimately related to materials science. We use path independence to help calculate various different properties of materials that we work with.

3. The path independence of potential is the property that allows circuits and batteries to function the way they do. Of course, this means that path independence is the entire reason that devices dependent on electric circuits, like the one you're reading this page on, operate!

History

Around 1800 an Italian doctor named Luigi Galvani found that touching a frog's leg to two different metals caused it to twitch. Alessandro Volta, a contemporary and rival or Galvani, studied these findings and concluded that a kind of electrical potential difference between the two metals caused a charge to flow through the frog's leg, firing the muscles to ultimately create a post-mortem twitch.

Volta found that, in the presence of significant electrical potential between two metals, electrical charge can flow through a metal wire (and through frog legs, salinated brine, etc). The analogy of the time was that current flowed through wire similarly to water in a pipe. Because of this discovery, Volta lives on through the concept of voltage and the associated unit of measurement - the volt.

References

Matter & Interactions Vol. II by Ruth W. Chabay and Bruce A. Sherwood

Chapter 16 Webassign

http://scienceline.ucsb.edu/getkey.php?key=4026

http://tutorial.math.lamar.edu/Classes/CalcIII/FundThmLineIntegrals.aspx

@@ Line 1: / Line 1: @@
-''Claimed by Josh Whitley Spring 2016, Edited by Emilie Pourchet Fall 2016''
+''Edited by Josh Whitley Spring 2016, Emilie Pourchet Fall 2016, William Nute Fall 2017''
-The concept of potential difference relies on what we call "Path Independence". This means that the path taken between the two observed locations does not affect the potential difference between the two locations. Therefore, any path can be taken to calculate the potential difference between two points. This can be very useful in some calculations, as will be shown later.
+'''CLAIMED BY WILLIAM NUTE FALL 2017'''
-==The Main Idea==
+See the picture drawn below. It is a clear example of path independence. If someone were to travel from point A to C to D to B, it would essentially be the same as traveling straight from A to B when we think about potential.
-=What is Electric Potential?=
-In order to understand why the electric potential difference is path independent, it can be helpful to define electric potential.
-Electric potential is the amount of electric potential energy that a point charge would have if it was located at some point in space (point B in this wiki page is how I will refer to this point in space), relative to an arbitrary reference point (point A). It equals the  work done by a constant external force in carrying the charge from point A to point B. Note that this does mean moving a point charge around in space before returning it to its original location means both the work done and the potential equal 0. Electric potential is measured in volts or Joules/Coulomb. Electric potential at infinity is effectively 0.
-=What is Path Independence?=
+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.
-Electric potential, as stated above, is entirely independent of the path taken to get from an initial to final state. When presented with a complex path from one point to another, choosing the easiest path (a straight line connecting the two points) can simplify the problem while still yielding the correct answer as a result of this principle.
-Note that path independence holds true for all potential energy, not just electric potential. For that reason, a useful frame of reference is the path independence of physical objects' potential energy in a  gravitic field that is introduced in physics 1 - electric potential path independence functions the same way.
+==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.
-===A Mathematical Model===
+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.
-The general representation of electric potential is the following:
+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.
-<math> \triangle V = V_b-V_a </math>
+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
-This formula can be expanded by summing the different components of the change in position within electric field components to calculate the change in electric potential. This equation is useful for proving path independence as it can be applied to multi-step path in straight line increments, proving when compared to a direct path that the change in electric potential is the same.
+<div style="text-align: center;"><math>\Delta V = -\int\vec{E}\cdot d\vec{l} = -\vec{E}\cdot \Delta \vec{l}</math> for constant <math>\vec{E}</math></div>
-<math> \triangle V = -(E_x \triangle x, E_y \triangle y, E_z \triangle z) </math>
+When you expand out the dot product, you get an expression
-===A Computational Model===
+<div style="text-align: center;"><math>\Delta V = -\langle E_x\Delta x, E_y\Delta y, E_z\Delta z\rangle</math></div>
-[https://trinket.io/glowscript/8d8cbfc025]
+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?
-Click the above link to go to a GlowScript page that details Path Independence of Electric Potential. 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!
+==Path Independence==
+Now that we know some logical reasoning behind why potential has the quality of path independence, what is it?
-In this simulation, you can:
+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>V_{AB} = V_B - V_A</math> in the first place.
-) Click the display window to progress the simulation, which will print a change in electric potential below corresponding to the parameters of your electric field and the size of your movement. By default, the electric field is (-50,0,0) N/C and the steps are arbitrary and simply designed to show a progression from point A to B by the particle.
+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.
-) See how two different paths (one blue and one white) result in the same change in electric potential, regardless of their stark differences in path length.
+===Mathematical Model===
+In more formal notation, you could see something like this:
-''Note that if you have difficulty seeing the simulation or the printed values, you can navigate to the menu in the top left of the coding window (represented by three parallel bars) and engage a fullscreen mode.''
+<div style="text-align: center;"><math>V_{AB} = (V_B - V_1) + (V_1 - V_2) + ... + (V_{n-1} - V_n) + (V_n - V_A)</math></div>
-''Note also that the default code is set to use an electric field with only an x component, for simplicity. Changes in the size of the movement steps are not supported.''
+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.
+===Computational Model===
+[https://trinket.io/glowscript/7bb82296bc 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=
+===Example 1===
-Below is an example that demonstrates the path independence of electric potential difference.
+This a basic example of using path independence to simplify the calculation of potential difference.
-[[File:phys1.jpg]]  [[File:Phys2.jpg]]
+<div style="text-align: center;">[[File:path-indep1.png|frameless|upright=3.5]] [[File:path-indep2.png|frameless|upright=3.5]]</div>
-Let's calculate the potential difference between A and B in the first scenario. Assume that the height difference between A and B is y and the x component increases by x.
+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> \Delta V = V_B - V_A </math>. We're given the field in between the plates is <math>\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> x </math> and <math> y </math> directions. This is an example of splitting up the path into the dot product, as discussed above.
-<math> \triangle V= -(E_x \triangle x, E_y \triangle y, E_z \triangle z) </math>
+Using path independence, we can get the equation <math> 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.
-<math> \triangle V= -E_x(x_1-0) + 0(y_1-0) + 0*0 </math>
+'''From A to C:'''
-<math> \triangle V = -E_xx_1 </math>
+<math> \Delta V = -E_xl  </math>
+<math> \Delta V = -Es </math>
-In the second scenario, let's divide this process into two parts to follow the path outlined from A to C and from C to B.
+'''From C to B:'''
-'''From A to C:'''
+<math> \Delta V = -E_yl </math>
+<math> \Delta V = 0 </math>
+To calculate total potential difference, we'll just add our two values together.
+<math> \Delta V=V_B - V_A = (V_B - V_C) + (V_C - V_A) </math>
+<math> \Delta V = -Es + 0 </math>
+<math> \Delta V = -Es </math>
+While in this example, the dot product calculation of <math>\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===
+[[File:Path-indep3.png|frameless|upright=3|center]]
+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> d </math> away from the center of the dipole. Each charge in the dipole has magnitude <math> q </math> with separation distance <math> 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> \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.
+[[File:Path-indep4.png|frameless|upright=3|center]]
+Our equation for this path is going to look something like the following:
+<div style="text-align: center;"><math> \Delta V = V_B - V_A = (V_B - V_\infty) + (V_\infty - V_A)</math></div>
+Since the field for a dipole varies with distance, we're going to have to integrate the field formula.
-<math> \triangle V = -(E_x \triangle x, E_y \triangle y, E_z \triangle z)  </math>
+<div style="text-align: center;"><math> \Delta V = -\int_{\infty}^{B}\vec{E}\cdot d\vec{l} - \int_{A}^{\infty}\vec{E}\cdot d\vec{l} </math></div>
-<math> \triangle V = -E_x(x_1-0)+ 0*0 + 0*0 </math>
+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> \triangle V = -E_xx_1</math>
+<div style="text-align: center;"><math> \Delta V = -\int_{A}^{\infty}\vec{E}\cdot d\vec{l} </math>
-'''From C to B:'''
+<math> \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> \triangle V = -(E_x \triangle x, E_y \triangle y, E_z \triangle z) </math>
+<math> \Delta V = -\frac{qs}{2\pi\epsilon_0}\int_{d}^{\infty} \frac{1}{x^3}dx </math>
-<math> \triangle V = -E_x*0 + 0*(y-0) + 0*0 </math>
+<math> \Delta V = \frac{qs}{2\pi\epsilon_0}\bigg(\frac{1}{2x^2}\bigg)\bigg|_{d}^{\infty} </math>
-<math> \triangle V = 0 </math>
+<math> \Delta V = \frac{qs}{2\pi\epsilon_0}(0) - \frac{qs}{2\pi\epsilon_0}\bigg(\frac{1}{2d^2}\bigg) </math>
-Now let's calculate the TOTAL potential difference for this scenario, it is the sum of the potential differences from A to B and from B to C.
+<math> \Delta V = -\frac{qs}{4\pi\epsilon_0d^2} </math></div>
-<math> \triangle V_{A to B} = \triangle V_{A to C} + \triangle V_{C to B} </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.
-<math> \triangle V_{total} = -E_xx_1 + 0 = -E_xx_1 </math>
+<div style="text-align: center;"><math> \int_{a}^{b}\vec{f}\cdot d\vec{r} = F(\vec{r}(b)) - F(\vec{r}(a)) </math> where <math> \nabla F = f </math></div>
-We can see that the results for both scenarios are the same, therefore the electric potential difference is independent of the path taken to calculate it.
+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.
-=Example 2=
+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:
-[[File:phys3.jpg]]
+<div style="text-align: center;"><math> -\int_{a}^{b}\vec{E}\cdot d\vec{l} = V(B) - V(A) </math></div>
-The above problem deals with a triple-plated, double-spaced, and parallel-plated capacitor and the associated potential difference as you cross over the center plate. Note that, intuitively, the electric fields are only multiplied by the portion of the path that falls within their respective capacitor space. They are then summed. This problem also deals with potential difference when path is reversed.
+Hopefully the above should look familiar - it's how we've been defining potential difference ever since we started learning it!
-In order to make this problem demonstrate Path Independence specifically, simply take a new path from A to B (by creating points in between) and calculate the displacement components times the corresponding electric field component of those mid points instead of the direct A to B path. You will find that all Y and Z components equal 0, which leaves only the X component of the path to be calculated. This will inevitably end up with the same potential difference as what is done above.
+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.
 ==Connectedness==
-''How is this topic connected to something that you are interested in? How is it connected to your major?''
+. Path independence is a topic that originally stems from multivariable calculus, which has always been one of my favorite subjects in mathematics. I loved studying the connection of so many different topics to culminate in concepts like Green's Theorem, Stoke's Theorem, and as mentioned in this page, the Fundamental Theorem of Line Integrals. Multivariable calculus has a huge connection to electromagnetism, and especially Maxwell's Equations.
-Spontaneous redox reactions often cause a potential disparity that can drive batteries. Electrochemistry, a subset of chemistry that I am interested in as an aspiring chemical engineer, spends a significant amount of time focusing on these interactions and the electrical output of different reaction combinations. In a sense, understanding the physics side of electrical potential helps round out my understanding of electrochemistry.
-''Is there an interesting industrial application?''
+.Thermodynamics and electric potential are both topics that rely greatly on path independence and are intimately related to materials science. We use path independence to help calculate various different properties of materials that we work with.
-In terms of strictly applying the idea of path independence, I would argue there is no industrial application. Though I would argue that related fields, such as electrochemistry, have sweeping industrial applications. The optimization of batteries relies partially on finding half reaction combinations that yield the largest positive potential. It's just that strictly dealing with path independence doesn't leave much room for innovation or invention - you need to involve related ideas. Perhaps the pathing of chemicals inside of a battery is made moot by this principle.
+. The path independence of potential is the property that allows circuits and batteries to function the way they do. Of course, this means that path independence is the entire reason that devices dependent on electric circuits, like the one you're reading this page on, operate!
 ==History==
 Around 1800 an Italian doctor named Luigi Galvani found that touching a frog's leg to two different metals caused it to twitch. Alessandro Volta, a contemporary and rival or Galvani, studied these findings and concluded that a kind of electrical potential difference between the two metals caused a charge to flow through the frog's leg, firing the muscles to ultimately create a post-mortem twitch.
 Volta found that, in the presence of significant electrical potential between two metals, electrical charge can flow through a metal wire (and through frog legs, salinated brine, etc). The analogy of the time was that current flowed through wire similarly to water in a pipe. Because of this discovery, Volta lives on through the concept of voltage and the associated unit of measurement - the volt.
-== See also ==
+==See also==
+===External Links===
 https://en.wikipedia.org/wiki/Voltage#See_also
@@ Line 122: / Line 157: @@
 ==References==
-Matter & Interactions volume II by Ruth W. Chabay and Bruce A. Sherwood.
+Matter & Interactions Vol. II by Ruth W. Chabay and Bruce A. Sherwood
-Chapter 16 Webassign Review
+Chapter 16 Webassign
 http://scienceline.ucsb.edu/getkey.php?key=4026
+http://tutorial.math.lamar.edu/Classes/CalcIII/FundThmLineIntegrals.aspx
 [[Category:Energy]]