thin rod

[logistic_guy]

A thin rod of length $\displaystyle 2L$ is centered on the $\displaystyle x$ axis. The rod carries a uniformly distributed charge $\displaystyle Q$. Determine the potential $\displaystyle V$ as a function of $\displaystyle y$ for points along the $\displaystyle y$ axis. Let $\displaystyle V = 0$ at infinity.

 

[khansaheb]

A thin rod of length $\displaystyle 2L$ is centered on the $\displaystyle x$ axis. The rod carries a uniformly distributed charge $\displaystyle Q$. Determine the potential $\displaystyle V$ as a function of $\displaystyle y$ for points along the $\displaystyle y$ axis. Let $\displaystyle V = 0$ at infinity.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this problem

 

[logistic_guy]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem

Thank you.

 

[logistic_guy]

$\displaystyle V = \int dV = \int k\frac{1}{r} \ dq$

 

[khansaheb]

$\displaystyle V = \int dV = \int k\frac{1}{r} \ dq$

What are 'r' ,'q' and 'k'?

 

[logistic_guy]

What are 'r' ,'q' and 'k'?

I think that they are variables related to the Potential Theorem.

 

[logistic_guy]

From accumulated experience when the charge $\displaystyle Q$ is uniformly distributed then the density per unit length is:

$\displaystyle \lambda = \frac{dQ}{dx}$

I am assuming that I placed the thin rod on a coordinate system! Then the integral becomes:

$\displaystyle V = \int k\frac{1}{r} \ dQ = \lambda k \int \frac{1}{r} \ dx$

It is still tough but I made a progress.

 

[logistic_guy]

With a sketch, I figured out how to write the variable $\displaystyle r$ in terms of the variable $\displaystyle x$. I am just lazy to show it here!

$\displaystyle V = \lambda k \int \frac{1}{r} \ dx = \lambda k \int \frac{1}{\sqrt{x^2 + y^2}} \ dx$

 

[logistic_guy]

$\displaystyle V = \lambda k \int_{-L}^{L} \frac{1}{\sqrt{x^2 + y^2}} \ dx = 2\lambda k \int_{0}^{L} \frac{1}{\sqrt{x^2 + y^2}} \ dx$

Ahhh this is an overwhelming integral

$\displaystyle x = yu$
$\displaystyle dx = y \ du$

$\displaystyle V = 2\lambda k \int_{0}^{\frac{L}{y}} \frac{1}{\sqrt{u^2 + 1}} \ du$

$\displaystyle u = \tan v$
$\displaystyle du = \sec^2 v \ dv$

$\displaystyle V = 2\lambda k \int_{0}^{\tan^{-1}\frac{L}{y}} \sec v \ dv = 2\lambda k \int_{0}^{\tan^{-1}\frac{L}{y}} \frac{\sec^2 v + \sec v \tan v}{\sec v + \tan v} \ dv$

$\displaystyle w = \sec v + \tan v$
$\displaystyle dw = \sec v \tan v + \sec^2 v \ dv$

$\displaystyle V = 2\lambda k \int_{v = 0}^{v = \tan^{-1}\frac{L}{y}} \frac{1}{w} \ dw = 2\lambda k\ln w \bigg|_{v=0}^{v = \tan^{-1}\frac{L}{y}}$


$\displaystyle = 2\lambda k\ln (\sec v + \tan v) \bigg|_{v=0}^{v = \tan^{-1}\frac{L}{y}}$


$\displaystyle = 2\lambda k \left[\ln\left(\sec \tan^{-1}\frac{L}{y} + \tan \tan^{-1}\frac{L}{y}\right) - \ln(\sec 0 - \tan 0)\right]$


$\displaystyle = 2\lambda k\left[\ln \left(\frac{\sqrt{L^2 + y^2}}{y} + \frac{L}{y}\right) - \ln 1\right]$


$\displaystyle = 2\lambda k\ln \left(\frac{\sqrt{L^2 + y^2}}{y} + \frac{L}{y}\right)$


$\displaystyle = \frac{kQ}{L}\ln \left(\frac{\sqrt{L^2 + y^2} + L}{y} \right)$