Potential Difference Problem - setting up the integral

[Taulant Sholla]

Homework Statement

Homework Equations

V=kq/x​

The Attempt at a Solution

I know the correct solution. It's...​

On my first attempt, rather than use (d+x) in the denominator and integrate from 0 to L, I instead used (x) and integrated from (d) to (L+d).

This produces the wrong answer, but why does it produce the wrong answer? Both approaches seem to capture the setup and handle the integration correctly. Obviously I'm wrong, but - again - why/how am I wrong? Thank you.

 

[BvU]

$\lambda$ is no longer $cdx$ if you change variables. You can avoid this kind of oversight by using a different integration variable name, e.g. $u$ or $v$, but not the same name $x$

 

[Taulant Sholla]

Thank you! However, question: where did I "change variables" and/or why is λ no longer cdx?

 

[BvU]

You changed the integration variable $x$ that runs from 0 to L into a variable with name e.g. $u$ that runs from d to d + L. In other words $u = x+d$. When $x = 0$, $\lambda = cx=0$, but when $u=d$, then $\lambda \ne cu$.

 

[Taulant Sholla]

Ah, thank you so much. This is very helpful. I appreciate your explanation very much!

 

[BvU]

You're welcome. That's what PF is for...