Sign of Potential Difference

Claimed by Tyler Quill, Spring 2016

This page provides an explanation to determine the sign of potential difference in which the sign shows whether energy is lost or gained by a moving charged particle.

The Main Idea

The potential difference is a quantity that represents a change in electric potential energy that a particle would experience moving through a region. Its most important quality is that it is independent of the charge in question. Multiplying the potential difference by the charge would yield the potential energy change said particle experiences. This relation is mathematically given by: [math]\displaystyle{ \Delta U }[/math] = [math]\displaystyle{ \Delta V\cdot q }[/math]

Another useful form of potential difference can be represented by [math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math]. By determining the direction of particle's path relative to the direction of electric field, the sign of potential difference can then be determined. The sign of the potential difference then shows if there is an increase or a decrease in potential energy, and a consequent change in the kinetic energy.

Note that both of these explanations for Potential Difference are valid and are useful in different scenarios.

An Energetic Approach

The total energy of a closed, insulated system (Just fancy thermodynamic terms that mean there is no exchange of particles with the surroundings and there is no heat flow) can be given by the following equation: [math]\displaystyle{ \Delta K }[/math] + [math]\displaystyle{ \Delta U }[/math] = [math]\displaystyle{ \Delta W }[/math], where [math]\displaystyle{ \Delta K }[/math] is the change in kinetic energy, [math]\displaystyle{ \Delta U }[/math] is the change in potential energy, and [math]\displaystyle{ \Delta W }[/math] is the amount of work done. In just about every scenario in this course, the work done will be equal to zero, so the we are left with: [math]\displaystyle{ \Delta K }[/math] + [math]\displaystyle{ \Delta U }[/math] = 0 This equation simply states that a decrease in the kinetic energy of a particle means an equal and opposite increase in the potential energy of the particle (energy is conserved). Using this relationship will allow one to better understand more complex problems of potential difference with moving particles and energy changes.

Example: A proton moves through a region of uniform electric field from point A to point B. While doing so, the proton's speed increases from 2,000 m/s to 4,300 m/s. Determine the potential difference between the proton's initial position and final position, and calculate the change in potential energy that an electron would experience moving through the same path.

1) Find the change in kinetic energy the proton experiences: [math]\displaystyle{ \Delta KE }[/math] = [math]\displaystyle{ KE_{final} - KE_{initial} }[/math], and since [math]\displaystyle{ KE }[/math]= [math]\displaystyle{ 1 \over 2 }[/math] [math]\displaystyle{ \cdot m \cdot (v^2) }[/math], we have:

[math]\displaystyle{ \Delta KE }[/math] =[math]\displaystyle{ 1 \over 2 }[/math] [math]\displaystyle{ \cdot m \cdot (v_{final}^2 -v_{initial}^2) }[/math]. Plugging in the numbers from the problem give: [math]\displaystyle{ \Delta KE }[/math] = [math]\displaystyle{ 1.210 \cdot 10^{-20} J }[/math] (recall that the mass of a proton is [math]\displaystyle{ 1.67 \cdot 10^{-27} Kg }[/math])

2) This change is equal and opposite of the change in potential energy of the proton. Thus the potential energy change is equal to [math]\displaystyle{ -1.210 \cdot 10^{-20} J }[/math]

3) Rearranging [math]\displaystyle{ \Delta U }[/math] = [math]\displaystyle{ \Delta V\cdot q }[/math], we find that the potential difference [math]\displaystyle{ \Delta V_{A \rightarrow B} }[/math] = [math]\displaystyle{ \Delta U \over q_{proton} }[/math]. This gives us a potential difference ( [math]\displaystyle{ \Delta V_{A \rightarrow B} }[/math] ) of [math]\displaystyle{ -0.07562 }[/math] Volts

4) We now have the potential difference, so calculating the change in potential energy that an electron would experience moving from A to B is simple. We know the following relationship [math]\displaystyle{ \Delta U }[/math] = [math]\displaystyle{ \Delta V\cdot q }[/math] and have already solved for the value of [math]\displaystyle{ \Delta V_{A \rightarrow B} }[/math], and the charge of an electron is known. Thus, multiplying these quantities gives us a [math]\displaystyle{ \Delta U_{A \rightarrow B} = 1.210 \cdot 10^{-20}J }[/math]

The general procedure for this type of problem is to calculate the change in potential energy and then use that to determine the potential difference based on the sign, and magnitude of the charge. It is also important to recognize how despite the potential energy changes for the proton and electron were of opposite sign, the sign of the potential difference is only positive. The potential difference does not care about the sign of the charge in question.

A Computational Model

Electric field lines always point from a region of high potential to a region of low potential. If given a diagram of a problem, draw the electric field lines and the sign of the potential difference can be found by [math]\displaystyle{ final - initial }[/math].

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As you can see in the diagram, the electric field lines point from A to B. Electric field lines point from regions on high potential to low potential. This means that point A has a higher potential than point B. Since [math]\displaystyle{ \Delta V = V_{final} - V_{initial} }[/math], then [math]\displaystyle{ \Delta V_{A \rightarrow B} = V_B -V_A }[/math]. We know that [math]\displaystyle{ V_{A} }[/math] is a greater value than [math]\displaystyle{ V_{B} }[/math], so their difference would give a negative value.

Note that for this scenario, the same methodology and logic can be used to determine the sign of the potential difference of points A and B. However, this scenario also includes point C, which is parrallel to point B relative to its distance along the electric field lines. Thus, neither C or B has a higher potential than the other, and their potential difference is consequently zero.

A Mathematical Model

Potential difference is the product of the electric field [math]\displaystyle{ \vec{E} }[/math] and the relative path [math]\displaystyle{ \Delta x }[/math]:

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math]

Sign of [math]\displaystyle{ \Delta V }[/math]

[math]\displaystyle{ \Delta x }[/math] in the direction of [math]\displaystyle{ \vec{E} }[/math]: negative

[math]\displaystyle{ \Delta x }[/math] in the opposite direction of [math]\displaystyle{ \vec{E} }[/math]: positvie

[math]\displaystyle{ \Delta x }[/math] is perpendicular to the direction of [math]\displaystyle{ \vec{E} }[/math]: [math]\displaystyle{ \Delta V }[/math]=0

Examples

Simple

If [math]\displaystyle{ x_i }[/math] = <3,0,0> m, [math]\displaystyle{ x_f }[/math] = <5,0,0> m, and [math]\displaystyle{ \vec{E} }[/math] = <100,0,0> V/m:

(Path is in the same direction as the electric field.)

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <5,0,0> - <3,0,0> = <2,0,0> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,0,0>●<2,0,0> = -200 V

If [math]\displaystyle{ x_i }[/math] = <5,0,0> m, [math]\displaystyle{ x_f }[/math] = <3,0,0> m, and [math]\displaystyle{ \vec{E} }[/math] = <100,0,0> V/m:

(Path is in the opposite direction of the electric field.)

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <3,0,0> - <5,0,0> = <-2,0,0> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,0,0>●<-2,0,0> = 200 V

If [math]\displaystyle{ x_i }[/math] = <3,0,0> m, [math]\displaystyle{ x_f }[/math] = <5,0,0> m, and [math]\displaystyle{ \vec{E} }[/math] = <0,100,0> V/m:

(Path is perpendicular to the electric field.)

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <5,0,0> - <3,0,0> = <2,0,0> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,0,0>●<0,2,0> = 0 V

Middling

If Location A = <3,0,0>, Location B = <5,-3,1> and E = <100, 100, 0>

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <5,-3,1> - <3,0,0> = <2,-3,1> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,100,0>●<2,-3,1> = <-200,300,0> V

In x-direction, there is an electric field in the same direction as the path, so the potential difference is negative.

In y-direction, there is an electric field in the opposite direction of the path, so the potential difference is positive.

In z-direction, since the electric field is perpendicular to the path, so the potential difference is zero.

Difficult

An electron starts from rest near one plate of a charged capacitor, and travels in the -x direction, passing through a tiny hole in the capacitor. At the instant shown in the diagram, the electron is at the origin. At this moment the magnetic field at location A, due to the electron, is out of the page. The left plate of the capacitor is positive, and the right plate is positive. What would be the potential difference from location D to location C? What is the sign of it?

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ \Delta V_C - \Delta V_D }[/math]

By conservation of energy, potential energy = kinetic energy

[math]\displaystyle{ q (\Delta V_C - \Delta V_D) }[/math] = [math]\displaystyle{ 1/2mv^2 }[/math], where [math]\displaystyle{ q }[/math] is the charge of electron, [math]\displaystyle{ m }[/math] is the mass of electron, and [math]\displaystyle{ v }[/math] is the velocity that the electron is traveling.

[math]\displaystyle{ \Delta V_C - \Delta V_D }[/math] = [math]\displaystyle{ 1/2mv^2/q }[/math]

The sign is positive since the electric field is going to the +x direction, whereas the electron is moving to the -x direction.

Connectedness

-Potential difference is connected to the concept of transmembrane potential which I am interested in. Potential difference between the outside and the inside of the cell membrane acts as a battery and provides important functions for us. The ion channels and ion pump proteins that are imbedded in the membrane allow ions to move across the membrane and create concentration gradients, which then create a potential difference that provides power to allow the transmission of the electric signals, such as those in our neurons and muscle cells. By opening and closing the ion channel, the signal is passed down to the next channels due to the change in potential. A brief video on this can be found here.

-As a Biomedical Engineering major, the concept of potential difference can be applied to the study of electric stimulation of cells. By causing the change in potential across the cells, voltage-dependent ion channels can be affected. From this concept, we can develop devices that are responsible for the signals of cells, such as a defibrillator.

-Industrial applications based on potential difference are again the devices that are used to affect the ion channels in which to allow activation or recovery of cell signals.

History

Most countries in the world and Europe use a voltage from 220 volts to 240 volts. In the other hand, most countries in the Americas and Japan use a voltage from 100 volts to 127 volts. In 19th century, Nikola Tesla determined that 60 Hz was the best frequency and preferred 240 volts for AC power while Thomas Edison preferred 110 volts. AEG, a German company, decided to build the first European generating facility with 50 Hz because 60 was dissatisfied by the sequence of the metric standard unit, and this standard was then spread out. Europe used to use 120 volts, but it was determined that higher voltage should be used to "get more power with less losses and voltage drop from the same copper wire diameter." The U.S did not end up changing because the cost for all the replacement was too high; fridge and washing-machine were already common in an average U.S. household in the 50s-60s but not in Europe. Then, problems like light bulbs burning out quickly led to the splitting voltage into two 120 volts.

See also

-Kinetic Energy and Potential Energy: The sign of potential difference shows whether energy is gained or lost from a moving particle. When there is an increase in potential energy [math]\displaystyle{ q \Delta V }[/math], there is a decrease in kinetic energy, and vice-versa.

-Polarization and Electric Field: Potential difference can be caused by polarization in which the charged particles move in different paths affect each other.

-Simple Circuits: Topics under simple circuits are mostly related to potential difference since batteries are involved, such as in RC circuit, Loop rule, and current.

Further reading

electric potential and capacitance

Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B.Sherwood (John Wiley & Sons 2015) (pp.635-637)

Dorf, Richard C. Sensors, Nanoscience, Biomedical Engineering, and Instruments. 3rd ed. N.p.: Taylor & Francis, 2006. Print.

External links

Optical measurements of transmembrane potential changes during electric field stimulation of ventricular cells.

Optical transmembrane potential measurements during defibrillation-strength shocks in perfused rabbit hearts.

Transmembrane Potential

References

G.W. Castellan, Physical Chemistry, 3rd Ed., Addison-Wesley, Reading,MA, 1983

Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B.Sherwood (John Wiley & Sons 2015) (pp.635-637)

SB1, Knisley, Blitchington TF, Hill BC, Grant AO, Smith WM, Pilkington TC, and Ideker RE. "Result Filters." National Center for Biotechnology Information. U.S. National Library of Medicine, Feb. 1993. Web. 22 Nov. 2015.

potential-difference-in-uniform-electric-field. Kshitij Education India. 22 Nov. 2015.

"Why Isn't There A standard Voltage around the World?" World Standards, 16 Aug. 2015. Web. 23 Nov. 2015.