Potential Difference at One Location

Written by Alex George

The Main Idea

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

A Mathematical Model

Potential at One Location

[math]\displaystyle{ V_{a} = V_{a} - V_{\infty } }[/math] where [math]\displaystyle{ V_{a} }[/math] is the potential location of interest and [math]\displaystyle{ V_{\infty } }[/math] is the potential at a location infinitely far away.

This equation is only valid when [math]\displaystyle{ V_{\infty } }[/math] is equal to zero.

Potential near a Point Charge

Parameters: A point charge with charge [math]\displaystyle{ q }[/math]. We are a distance [math]\displaystyle{ x }[/math] from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

[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} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} }[/math]

The sign of the potential depends on the sign of the charge on the particle:

if [math]\displaystyle{ q \lt 0, V_{x} \lt 0 }[/math]

if [math]\displaystyle{ q \gt 0, V_{x} \gt 0 }[/math]

Potential Energy at a Single Location

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

[math]\displaystyle{ U_{A} = qV_{A} }[/math]

Substituting in the derived equation from the previous section for [math]\displaystyle{ V_{A} }[/math], we obtain the following:

[math]\displaystyle{ U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) }[/math] where [math]\displaystyle{ x }[/math] is the separation between the two particles, and [math]\displaystyle{ q_{1} }[/math] and [math]\displaystyle{ q_{2} }[/math] are the respective charges of the particles.

A Visual Model

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

Figure 1. Potential vs. Distance for a Point charge with Charge [math]\displaystyle{ e }[/math]

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

[math]\displaystyle{ E_{x} = -\frac{\partial V}{\partial x} }[/math]

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]

Connectedness

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See also

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Further reading

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External links

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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.