Potential Difference at One Location: Difference between revisions
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<math>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> | <math>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> q < 0, V_{x} < 0 </math> | |||
if <math> q > 0, V_{x} > 0 </math> | |||
===A Computational Model=== | ===A Computational Model=== | ||
Revision as of 13:38, 5 December 2015
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]
A Computational Model
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