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The next day, your physics teacher brings in a Van der Graaf generator, which causes your classmate's hair to stand on end when they touch it. You are excited to impress your teacher and raise your hand enthusiastically when a new volunteer is required. After your hair is made to also stand on end, to your classmates' amusement, you hear the magic words from your teacher, "Well done! You have demonstrated potential." No sweeter sound could have entered your ears and you return to your seat satisfied that you have atoned for any physics-related sins.
If only you knew that the potential that your teacher was actually referring to was electric potential... In this article, we will discuss the electric potential due to a point charge, so that you may never make this mistake again.
Defining the Electric Potential due to a Point Charge
We know that, in reality, charged particles like protons and ions have a definite size and occupy some volume in space. It may be a tiny value but it does exist. For ease of understanding in this article, we are going to assume that all charges only occupy a single point in space. We will refer to these objects as point charges. We know that any charged particle will have an electric field, which is no different for point charges. The electric field lines for point charges are radial and point into or away from the charge (depending on the sign of the charge). We need to define a new quantity, the electric potential, and we will do so for a point charge specifically.
The electric potential \(V\) at a point in the electric field of a point charge is the work done \(W\) per unit positive charge \(q\) in bringing a small test charge from infinity to that point.
Simply we can write this mathematically as \[V=\frac{W}{q}.\] Adjacent points that have equal electric potential form lines of equipotential, also called isolines.
The Formula of Electric Potential due to a Point Charge
If two points lie on the same isoline, no work is done in moving a charged particle between those points. The isolines produced by point charges form concentric circles centered on the charge. It is clear that the potential \(V\) is related to the distance \(r\) from the charge \(q\). In fact, \[V=\frac{1}{4\pi \varepsilon_0}\frac{q}{r},\] where \(\varepsilon_0\) is a constant known as the permittivity of free space and has the value \(\varepsilon_0 = 8.85\times10^{-12}\,\mathrm{F\,m^{-1}}.\) The SI unit of measurement of potential is the \(\text{volt, V,}\) which is equivalent to the \(\text{joule-per-coulomb, } \mathrm{J\,C^{-1}}.\) A graph of due potential against distance due to a positive charge and due to a negative charge is shown in Fig. 1 below.
The graph takes on a hyperbolic shape representing the drop in potential as distance increases. It is flipped about the distance axis for a negative charge. This can be seen from the mathematical expressions, firstly for a positive charge, \[V_{+}=\frac{1}{4\pi \varepsilon_0}\frac{+q}{r},\] and then for a negative charge, \[V_{-}=\frac{1}{4\pi \varepsilon_0}\frac{-q}{r}.\]
We can also relate the electric potential to the average magnitude of the electric field \(\left|\vec{E}\right|\) as follows, \[\left|\vec{E}\right|=\left|\frac{\Delta V}{\Delta r}\right|.\] The average magnitude of the electric field between two points is equal to the magnitude of change in electric potential \(\Delta V\) divided by the change in position between those points \(\Delta r\) in the field. The change in potential \(\Delta V\) between two points is also called the potential difference between those points.
Derivation of the Formula for the Electric Potential due to a Point Charge
We can derive the equation above by considering the example of two positive charges \(q\) and \(Q\) separated by a distance \(r.\) This is represented in Fig. 2 below.
The force \(F_{qQ}\) that charge \(q\) exerts on \(Q\) is equal and opposite to the force \(F_{Qq}\) that charge \(Q\) exerts on \(q.\) We can call the magnitude of this force \(F.\) From Coulomb's law, \[F=\frac{1}{4\pi \varepsilon_0}\frac{qQ}{r^2},\] and the electric potential energy \(E_\mathrm{P}\) is the same as the work done \(W\) to bring two charges to points at which their separation is \(r,\) \[E_\mathrm{P}=W=\frac{1}{4\pi \varepsilon_0}\frac{qQ}{r}.\] The definition of electric potential tells us that the work done per unit charge in bringing charge \(Q\) from infinity to a distance \(r\) from charge \(q\) is given by \[\begin{align}V&=\frac{W}{Q}\\&=\frac{1}{\cancel{Q}} \cdot \frac{1}{4\pi \varepsilon_0} \frac{q\cancel{Q}}{r}\\&=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}, \end{align}\] which is the same as the first equation stated above.
Electric Potential due to a Point Charge Diagram
If we have a uniform electric field, we know that the electric field lines will be parallel to each other and point in the same direction. That direction will be determined by the sign of the charge on the surface of the object generating the potential. The equation for electric potential tells us that at different distances \(r\) from the surface, there will be different potentials. However, along a line that is parallel to the surface, the potential will be constant, as all points on that line are equidistant from the surface. These lines of constant potential are called isolines and for a uniform field, they appear as in Fig. 3 below.
Note that the isolines are always perpendicular to the field lines. This is always necessary since any component of the electric field along the direction of an isoline will cause an electric force on a charge along that line. Work would be done along that isoline and potential would not remain constant which cannot occur.
The scenario is different for a point charge. The field lines would be radial but we would require that the isolines always be perpendicular to them. The isolines would therefore form concentric circles centered on the point charge \(q.\) Fig. 4 below shows the field lines and isolines due to a positive point charge.
The circular isolines mean that the potential is constant along a circular path of radius \(r\) surrounding the point charge. If we think quite classically and assume that electrons orbit the nucleus of an atom in a circular path, this would be why the nucleus does not work on electrons.
Electric Potential due to a Point Charge: Examples
Now that we have seen how the electric potential of a point charge varies with distance, we can work our way through some examples relating to this concept.
Question: The electric potential energy between an electron and proton is \(9.6\times 10^{-17}\,\mathrm{J}.\) Calculate the electric potential of the electron at the position of the proton assuming that both can be treated as point charges.
Answer: Recall that the charge of a proton is \(1.60\times 10^{-19}\,\mathrm{C}.\) The electric potential \(V\) due to the electron at the position of the proton is the work done per unit charge in bringing the proton to that point in the electric field of the electron. \[\begin{align}V&=\frac{W}{Q}\\[4 pt]&=\frac{9.6\times 10^{-17}\,\mathrm{J}}{1.60\times 10^{-19}\,\mathrm{C}}\\[4 pt] &=600\,\mathrm{J\,C^{-1}}\\[4 pt]&=600\,\mathrm{V}. \end{align}\] The electric potential due to the electron at the position of the proton is \(600\,\mathrm{V}.\)
We can now move on to slightly more complex examples.
Question: Calculate the electric potential of a \(2.0\,\mathrm{nC}\) point charge at a distance of \(0.50\,\mathrm{cm}\) from the charge.
Answer: We can use the equation relating potential \(V\) to distance \(r,\) \[\begin{align} V&=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}\\[2 pt]&=\frac{1}{4\pi \left(8.85\times10^{-12}\,\mathrm{F\,m^{-1}}\right)}\left(\frac{2.0\times 10^{-9}\,\mathrm{C}}{0.50 \times 10^{-2}\,\mathrm{m}}\right)\\[4 pt]&=3\,600\,\mathrm{C\,F^{-1}}\\[4 pt]&=3\,600\,\mathrm{V}. \end{align}\] The electric potential of this charge is \(3\,600\,\mathrm{V}\), at a distance of \(0.50\,\mathrm{cm}\) from the charge.
Lastly, we can take a look at how a potential difference between two points affects the magnitude of the electric field in that region.
Question: Calculate the average magnitude of the electric field between two points which have a potential difference of \(150\,\mathrm{V}\) between them, and are separated by a distance of \(2.5\,\mathrm{cm}.\)
Answer: We can use the equation that relates the average magnitude of the electric field \(\left|\vec{E}\right|\) to the change in potential with position \(\left|\frac{\Delta V}{\Delta r}\right|,\) \[\begin{align} \left|\vec{E}\right|&=\left|\frac{\Delta V}{\Delta r}\right|\\[4 pt]&=\left|\frac{150\, \mathrm{V}}{2.5\times 10^{-2}\,\mathrm{m}}\right|\\[4 pt]&=6.0\times 10^{3}\,\mathrm{V\,m^{-1}}.\end{align}\] The electric field has an average value of \(6.0\times 10^{3}\,\mathrm{V\,m^{-1}}\) between the two points.
Electric Potential Due to a Point Charge - Key takeaways
- The electric potential \(V\) at a point in the electric field of a point charge is the work done \(W\) per unit positive charge \(q\) in bringing a small test charge from infinity to that point, \[V=\frac{W}{q}.\]
- For a point charge, the potential \(V\) is related to the distance \(r\) from the charge \(q\), \[V=\frac{1}{4\pi \varepsilon_0}\frac{q}{r}.\]
- The SI unit of measurement of potential is the \(\text{volt, V.}\)
- The average magnitude of the electric field \(\left|\vec{E}\right|\) between two points is equal to the magnitude of change in electric potential \(\Delta V\) divided by the change in position between those points \(\Delta r\) in the field, \[\left|\vec{E}\right|=\left|\frac{\Delta V}{\Delta r}\right|.\]
- Isolines are always perpendicular to field lines.
- No work is done by the electric force along an isoline.
- For a uniform field, field lines are parallel to each other and isolines are parallel to each other but perpendicular to the field lines.
- For the field of a point charge, field lines are radial and isolines form concentric circles centered on the charge.
References
- Fig. 1 - A graph of electric potential vs distance shows an inverse relationship for a positive charge and the curve is flipped about the distance axis for a negative charge. StudySmarter Originals
- Fig. 3 - The field lines for a uniform electric field are parallel to each other. The isolines or lines of equipotential are also parallel to each other but are perpendicular to the field lines at all times. StudySmarter Originals
- Fig. 4 - The field lines for the electric field of a point charge are radial. The isolines are always perpendicular to the field lines and so form concentric circles centered on the charge. StudySmarter Originals
- Fig. 2 - The electric force between two charges can be used to find the electric potential due to one of the charges, StudySmarter Originals
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Frequently Asked Questions about Electric Potential due to a Point Charge
What factors determine electric potential?
The magnitude of the charge creating the field, the distance from this charge, and the medium in which the charge exists.
What is the formula of electric field due to a point charge?
The electric field E of a point charge q, at a distance r from it, is given by E=kq/r^2.
What is the difference between charge and point charge?
Charge can be distributed over a large area but a point charge considers all of the charge to be located at a single point in space.
What is an example of electric potential due to a point charge?
The voltage of a battery is an example of the electric potential difference between its ends.
What is a point charge in physics?
Charge can be distributed over a large area but a point charge considers all of the charge to be located at a single point in space.
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