Potential Difference at One Location
Written by Kayden Less SPRING 26'
The Main Idea
Electric potential is the electric potential energy per unit charge at a point in space. It is a scalar quantity and provides an alternative to analyzing electric interactions using forces. Because electric potential is scalar, contributions from multiple charges may be added algebraically through superposition, often simplifying electrostatic problems.
Electric potential difference is defined as the negative work done per unit charge by the electric field as a test charge moves between two points:
[math]\displaystyle{ \Delta V = V_B - V_A = -\int_A^B \vec E \cdot d\vec s }[/math]
If the reference point is chosen to be infinitely far away and
[math]\displaystyle{ V_{\infty}=0 }[/math]
then electric potential at a point can be determined relative to infinity.
Potential near a Point Charge
Parameters: A point charge with charge [math]\displaystyle{ q }[/math]. A point is located a distance [math]\displaystyle{ r }[/math] from the charge.
Using Coulomb’s law,
[math]\displaystyle{ E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2} }[/math]
and the definition of potential,
[math]\displaystyle{ V_r=V_r-V_{\infty} =-\int_{\infty}^{r}\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}dr }[/math]
which gives
[math]\displaystyle{ V_r=\frac{1}{4\pi\epsilon_0}\frac{q}{r} }[/math]
This shows electric potential decreases with increasing distance and depends on the sign of the source charge.
if [math]\displaystyle{ q\gt 0,\quad V_r\gt 0 }[/math]
if [math]\displaystyle{ q\lt 0,\quad V_r\lt 0 }[/math]
Electric Potential and Electric Field
Electric potential is related to electric field through
[math]\displaystyle{ \vec E=-\nabla V }[/math]
which shows electric fields point in the direction of decreasing potential.
For a uniform electric field,
[math]\displaystyle{ \Delta V=-Ed }[/math]
when displacement is parallel to the field.
Superposition of Electric Potential
For several point charges, electric potential is found by summing contributions from each charge:
[math]\displaystyle{ V_{net}=\sum_i \frac{1}{4\pi\epsilon_0}\frac{q_i}{r_i} }[/math]
Because potential is scalar, these contributions add algebraically rather than vectorially.
Potential Energy at a Single Location
The electric potential energy of a charge [math]\displaystyle{ q }[/math] placed in potential [math]\displaystyle{ V }[/math] is
[math]\displaystyle{ U=qV }[/math]
This relationship connects electric potential directly to energy methods and is fundamental in analyzing charged particle motion, circuits, and capacitors.
Importance of Electric Potential
Electric potential is a foundational idea in electrostatics because it links work, energy, electric fields, voltage, and later topics such as capacitance and circuits. It provides a powerful method for solving problems where force-based analysis may be difficult.
Here is a video explaining the above concepts with more examples [1].
Examples
Here are a couple of problems that illustrate the concepts presented above.
Simple Examples
Example 1 - Potential at a Location Near a Sphere
What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?
Solution:
As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.
From the above derivation of the potential for a single point charge, we have:
[math]\displaystyle{ V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) }[/math]
[math]\displaystyle{ V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V }[/math]
Example 2 - Charge Required to Produce a Specific Potential
What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?
Solution
The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:
[math]\displaystyle{ V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) }[/math]
[math]\displaystyle{ q = {4\pi\varepsilon _{0}} x V_{x} }[/math]
[math]\displaystyle{ q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C }[/math]
Difficult Examples
Example 1 - Potential Along the Axis of a Ring
What is the potential a distance [math]\displaystyle{ z }[/math] from the center of the ring?
Solution:
The electric field of a ring is given by the following equation:
[math]\displaystyle{ E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} }[/math] where [math]\displaystyle{ z }[/math] is the distance from the center of the ring, [math]\displaystyle{ R }[/math] is radius of the ring, and [math]\displaystyle{ Q }[/math] is the charge on the ring.
This equation can be integrated with respect to [math]\displaystyle{ z }[/math] and the appropriate bounds to obtain the potential at a location [math]\displaystyle{ z }[/math] from the center of the ring:
[math]\displaystyle{ V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz }[/math]
[math]\displaystyle{ V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} }[/math]
Difficult Example 2 — Superposition of Potential
Parameters: Two point charges contribute to the electric potential at point [math]\displaystyle{ P }[/math].
[math]\displaystyle{ q_1=+5\mu C,\quad r_1=0.20m }[/math] [math]\displaystyle{ q_2=-3\mu C,\quad r_2=0.10m }[/math]
Using superposition,
[math]\displaystyle{ V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right) }[/math]
Substituting values,
[math]\displaystyle{ V=(9\times10^9)\left(\frac{5\times10^{-6}}{0.20}-\frac{3\times10^{-6}}{0.10}\right) }[/math] [math]\displaystyle{ V=(9\times10^9)(25\times10^{-6}-30\times10^{-6}) }[/math] [math]\displaystyle{ V=(9\times10^9)(-5\times10^{-6}) }[/math] [math]\displaystyle{ V=-4.5\times10^4V }[/math]
The negative result indicates the contribution from the negative charge dominates at point [math]\displaystyle{ P }[/math].
Computational Model
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Connectedness
- How is this topic connected to something that you are interested in? This topic is related to the concept of Potential Energy, or the capacity to do work in a system. Electric Potential is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.
- How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.
- Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.
Real-World Applications
Electric potential appears in many technologies and physical systems and serves as an important practical application of electrostatics.
Capacitors
Capacitors store electrical energy through a potential difference between two conductors.
Energy stored in a capacitor is
[math]\displaystyle{ U=\frac12CV^2 }[/math]
where [math]\displaystyle{ C }[/math] is capacitance and [math]\displaystyle{ V }[/math] is voltage.
Applications include:
Electronic circuits Camera flashes Energy storage devices Power conditioning systems
Particle Accelerators
Particle accelerators use electric potential differences to accelerate charged particles.
A particle accelerated through a potential difference gains kinetic energy according to
[math]\displaystyle{ \Delta K=q\Delta V }[/math]
Applications include:
Physics research Radiation therapy Materials science
Electrostatic Precipitators
Electrostatic precipitators use electric potential to create electric fields that remove charged pollutants from industrial exhaust streams.
Charged particles are attracted toward collection plates through electrostatic forces, reducing emissions.
Power Transmission
Power transmission systems rely on high voltages to reduce resistive losses during long-distance energy transfer.
Power is related by
[math]\displaystyle{ P=IV }[/math]
Increasing voltage allows lower current for the same power output, reducing losses:
[math]\displaystyle{ P_{loss}=I^2R }[/math]
This is why high-voltage transmission lines are used.
Electric potential is therefore not only a theoretical concept, but also a practical engineering tool used across science and engineering.
Common Mistakes
Students often make several common mistakes when working with electric potential.
Confusing electric potential and electric field.
Using
[math]\displaystyle{ V=\frac{kq}{r} }[/math]
instead of distinguishing it from
[math]\displaystyle{ E=\frac{kq}{r^2} }[/math] Forgetting electric potential is a scalar quantity and adds algebraically rather than vectorially. Ignoring the sign of the source charge when determining whether potential is positive or negative.
if [math]\displaystyle{ q\gt 0 }[/math], then [math]\displaystyle{ V\gt 0 }[/math]
if [math]\displaystyle{ q\lt 0 }[/math], then [math]\displaystyle{ V\lt 0 }[/math]
Mixing up electric potential [math]\displaystyle{ V }[/math]
and electric potential energy
[math]\displaystyle{ U=qV }[/math] Forgetting electric potential is path independent and depends only on endpoints.
Exam Tip
For multiple charges, use superposition:
[math]\displaystyle{ V_{net}=\sum_i \frac{kq_i}{r_i} }[/math]
Since potential is scalar, add potentials algebraically.
History
The SI unit for electric potential is the volt, which is named in honor of the Italian physicist Count Alessandro Volta (1745-1827), who invented the voltaic pile, the first chemical battery.
Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.
Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.
More information on Alessandro Volta can be found at Wikipedia [2] .
See also
Potential Difference in a Uniform Field
Potential Difference Path Independence
Potential Difference of point charge in a non-Uniform Field
Potential Difference in an Insulator
Further reading
Chabay, Ruth W. Matter and Interactions, Ch. 16.7. 4th ed. New York: Wiley, 2015. Print.
Urone, Paul Peter., et al. College Physics, Ch. 19.3. Houston, Texas: OpenStax College, Rice University, 2013. Print.
External links
Youtube Video on Electric Potential of a Point Charge
References
Chabay, Ruth W. Matter and Interactions. 4th ed. New York: Wiley, 2015. Print.
Urone, Paul Peter., et al. College Physics. Houston, Texas: OpenStax College, Rice University, 2013. Print.
Wikipedia.org