A rod of length L lies along the x-axis with the left end at the origin. It has non-uniform charge density given by [tex]\lambda = \alpha x[/tex]. where [tex]\alpha[/tex] is a positive constant. Calculate the potential at a position A, which is a distance d to the left of the origin.(adsbygoogle = window.adsbygoogle || []).push({});

Here is what i did:

we have [tex] dV = \frac{k_e dq}{r}[/tex]

then i made use of the [tex]\lambda = \alpha x[/tex]

and we have [tex]V=\int\frac{k_e\lambda dx}{x} = \int\frac{k_e\alpha x dx}{x} = k_e\alpha (d+L-d)[/tex] where my limits of integration were L to d+L in the second integral

however the answer provided says that [tex]V=k_e\alpha (L-d ln(1 + \frac{L}{d})[/tex].

where did i go wrong?

thanks

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# Homework Help: Potential due to a rod

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