# Integrating An Electric Field Equation

How does one go about integrating something like $$V = \int_{a}^{b}\frac{Q}{2\pi r\epsilon_{0}\epsilon_{r}}dr$$Where the values of $a$,$b$,$\epsilon$,$Q$ are given and V is the potential difference

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Are $\epsilon_0,\epsilon_r,Q$ (real) constants? – Andrew Aug 2 '12 at 14:27
$\int_{a}^{b}\frac{1}{r}dr=\ln \left\vert b\right\vert -\ln \left\vert a\right\vert +C$ – Américo Tavares Aug 2 '12 at 14:30
I should have written $\int_{a}^{b}\frac{1}{r}dr=\ln \left\vert b\right\vert -\ln \left\vert a\right\vert$ – Américo Tavares Aug 2 '12 at 17:31

We simply use standard calculus techniques. Remember, $\varepsilon_{0}$, $\varepsilon_{r}$, $2\pi$ and $Q$ are real-valued constants.
$$V=\int_{a}^{b}{\frac{Q}{2\pi r\varepsilon_{0}\varepsilon_{r}}\:dr}=\frac{Q}{2\pi\varepsilon_{0}\varepsilon_{r}}\int_{a}^{b}{\frac{1}{r}\:dr}$$
We also know that $\int{\frac{1}{r}\:dr}=\ln{|r|}+c_{1}$, so we have:
$$V=\frac{Q}{2\pi\varepsilon_{0}\varepsilon_{r}}\left(\ln|b|-\ln|a|\right)=\frac{Q}{2\pi\varepsilon_{0}\varepsilon_{r}}\ln{\left|\frac{b}{a}\right|}$$
Given that we can presume $r$ represents a (non-negative) radius, the modulus can be leaved apart. – enzotib Aug 2 '12 at 17:22