Sign of Potential Difference

Claimed by Wendy Sheu

This page provides an explanation to determine the sign of potential difference in which the sign shows whether energy is lost or gained by a moving charged particle.

The Main Idea

By determining the direction of path relative to the direction of electric field, the sign of potential difference can then be determined. The sign of potential difference then shows if there is an increase or a decrease in potential energy, as well as kinetic energy.

A Mathematical Model

Potential difference is the product of the electric field [math]\displaystyle{ \vec{E} }[/math] and the relative path [math]\displaystyle{ \Delta x }[/math]:

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math]

Sign of [math]\displaystyle{ \Delta V }[/math]

[math]\displaystyle{ \Delta x }[/math] in the direction of [math]\displaystyle{ \vec{E} }[/math]: negative

[math]\displaystyle{ \Delta x }[/math] in the opposite direction of [math]\displaystyle{ \vec{E} }[/math]: positvie

[math]\displaystyle{ \Delta x }[/math] is perpendicular to the direction of [math]\displaystyle{ \vec{E} }[/math]: [math]\displaystyle{ \Delta V }[/math]=0

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

Simple

If [math]\displaystyle{ x_i }[/math] = <3,0,0> m, [math]\displaystyle{ x_f }[/math] = <5,0,0> m, and [math]\displaystyle{ \vec{E} }[/math] = <100,0,0> V/m:

(Path is in the same direction as the electric field.)

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <5,0,0> - <3,0,0> = <2,0,0> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,0,0>●<2,0,0> = -200 V

If [math]\displaystyle{ x_i }[/math] = <5,0,0> m, [math]\displaystyle{ x_f }[/math] = <3,0,0> m, and [math]\displaystyle{ \vec{E} }[/math] = <100,0,0> V/m:

(Path is in the opposite direction of the electric field.)

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <3,0,0> - <5,0,0> = <-2,0,0> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,0,0>●<-2,0,0> = 200 V

If [math]\displaystyle{ x_i }[/math] = <3,0,0> m, [math]\displaystyle{ x_f }[/math] = <5,0,0> m, and [math]\displaystyle{ \vec{E} }[/math] = <0,100,0> V/m:

(Path is perpendicular to the electric field.)

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <5,0,0> - <3,0,0> = <2,0,0> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,0,0>●<0,2,0> = 0 V

Middling

If Location A = <3,0,0>, Location B = <5,-3,1> and E = <100, 100, 0>

[math]\displaystyle{ \Delta \vec{x} }[/math] = [math]\displaystyle{ x_f }[/math] - [math]\displaystyle{ x_i }[/math] = <5,-3,1> - <3,0,0> = <2,-3,1> m

[math]\displaystyle{ \Delta V }[/math] = -[math]\displaystyle{ \vec{E} }[/math]●[math]\displaystyle{ \Delta \vec{x} }[/math] = -<100,100,0>●<2,-3,1> = <-200,300,0> V

In x-direction, there is an electric field in the same direction as the path, so the potential difference is negative.

In y-direction, there is an electric field in the opposite direction of the path, so the potential difference is positive.

In z-direction, since the electric field is perpendicular to the path, so the potential difference is zero.

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