Potential due to a finite charged wire

[lorenz0]

Homework Statement

A wire of finite length has linear positive charge density $\lambda$
What is the potential at point O?

Relevant Equations

$V(r)=\frac{q}{4\pi\varepsilon_0 r}$

Considering a reference frame with $x=0$ at the leftmost point I have for the leftmost piece of wire: $\int_{x=0}^{x=2R}\frac{\lambda dx}{4\pi\varepsilon_0 (3R-x)}=\frac{\lambda ln(3)}{4\pi\varepsilon_0}$.
The potential at O due to the semicircular piece of wire at the center is $\int_{\theta=0}^{\theta=\pi}\frac{\lambda Rd\theta}{4\pi\varepsilon_0 R}=\frac{\lambda}{4\varepsilon_0}$.
The potential at O due to the rightmost piece of wire is, by symmetry, the same as that due to the leftmost piece of wire $(\int_{x=R}^{x=3R}\frac{\lambda dx}{4\pi\varepsilon_0 x}=\frac{\lambda ln(3)}{4\pi\varepsilon_0}).$

So, the total potential at O is $V(O)=2\frac{\lambda ln(3)}{4\pi\varepsilon_0}+\frac{\lambda}{4\varepsilon_0}=\frac{\lambda(\pi+ln(9))}{4\pi\varepsilon_0}$.

Does this make sense? Thanks

 

[BvU]

Wouldn't the left and right straight sections yield the same potential at O ? (you know, symmetry and all that...)

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[haruspex]

$(4R+x)$? How do you get that?

 

[lorenz0]

Wouldn't the left and right straight sections yield the same potential at O ? (you know, symmetry and all that...)

$\$

Yes; I have edited my answer, thanks.

 

[lorenz0]

That was a mistake, which I have corrected thanks to user BvU.