Potential Difference at One Location

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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 bet 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

1) 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{q}{\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} }[/math]


Difficult

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