Potential of a point from charged rod.

AI Thread Summary
The discussion revolves around the challenge of determining the potential from a charged rod, specifically regarding the integration limits when calculating the potential at a point on the x-axis. The user encounters two different results based on the limits chosen for integration, leading to confusion over which is correct. It is clarified that the discrepancy arises from the need to adjust the differential element when changing limits, specifically using d(-x) instead of dx. This highlights the importance of maintaining consistent limits and understanding the relationship between logarithmic functions in integration. The conversation concludes with an acknowledgment of the learning gained from the interaction.
Piyu
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Homework Statement


Hello, the problem i have with finding the potential is the integration limits. Assuming both the rod and point lies on the x axis. When i integrate the sum of many slices of rod, ill get 2 answers(negative and positive) depending on the limits i use.

Example i ended up integrating dx/x which means that if i set the far end of the rod as lower limit ill end up with ln(x/L+x) instead of ln((L+x)/X).

So the question is which of it is true and is there a standard way to approach this?


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The Attempt at a Solution

 
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Hi Piyu! :smile:
Piyu said:
Example i ended up integrating dx/x which means that if i set the far end of the rod as lower limit ill end up with ln(x/L+x) instead of ln((L+x)/X).

One is minus the other (logs of inverses) …

and the minus comes because you should have replaced dx by d(-x) if you went the other way! :wink:
 
AH! So whenever i integrate over a variable like dx, it has to go from negative to positive(the right way). Never knew that :P Thanks for the help appreciate it!
 
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