How come electric potentials have different signs?

AI Thread Summary
Electric potentials can have different signs depending on the direction of integration when calculating from a point charge. Integrating from R to infinity yields a positive potential, while integrating from infinity to R results in a negative potential due to the change in direction of the electric field and differential elements. This sign change is a consequence of the vector nature of electric fields and the limits of integration affecting the direction of the ds elements. Careful attention to these directions is crucial when performing the integration. Understanding this concept clarifies why electric potential can appear with different signs in different contexts.
ben123467
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Homework Statement


The textbook says to find the electric potential due to a point charge by moving a test charge from R to infinity, and integrating using the equation below. But when I integrate from infinity to R, the sign switches. Why is that? Both times, electric potential at V is 0.

This is confusing because to find the electric potential from point charge, you integrate from R to infinity, but to find the potential energy of a system, you integrate from infinity to R

Homework Equations


Rinf E dot ds = Vf - Vi

The Attempt at a Solution


From R to inf
Vi = K q/r

From inf to R
Vf = -K q/r
OwEc2Ur.jpg

https://i.imgur.com/OwEc2Ur.jpg in case not clear
 

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Hi ben123467,

:welcome:

You need to be a bit careful when you integrate an expression that comes from vectors. The electric field has a direction, and so do the ds elements. The ds elements have a direction that depends upon your limits of integration. When you exchange the limits, the direction of the ds elements change. In the scalar version of the integration, that's handled by the ##cos(\theta)## in the dot product.
 
gneill said:
Hi ben123467,

:welcome:

You need to be a bit careful when you integrate an expression that comes from vectors. The electric field has a direction, and so do the ds elements. The ds elements have a direction that depends upon your limits of integration. When you exchange the limits, the direction of the ds elements change. In the scalar version of the integration, that's handled by the ##cos(\theta)## in the dot product.

I took into account the opposite directions of E and ds though. I uploaded a picture, can you check my math?
 
Kind of difficult to read your image due to the scale. A closeup of the textbook page posted separately might help, and either the same for your handwritten work or, much preferred, type out your work so that helpers can easily quote individual lines in their replies. The ##\Sigma## icon in the edit window header bar gives you access to math symbols, and you can also use LaTeX syntax to render math expressions in a very nice way.
 
gneill said:
Kind of difficult to read your image due to the scale. A closeup of the textbook page posted separately might help, and either the same for your handwritten work or, much preferred, type out your work so that helpers can easily quote individual lines in their replies. The ##\Sigma## icon in the edit window header bar gives you access to math symbols, and you can also use LaTeX syntax to render math expressions in a very nice way.

There's a link to imgur below the image if that helps. You can zoom in from there.
 
ben123467 said:
There's a link to imgur below the image if that helps. You can zoom in from there.
Still no joy. The zoom is the same as what's available for your in-post image.
 
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