Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks in advance.

share|improve this question
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

1 Answer 1

up vote 3 down vote accepted

We simply use standard calculus techniques. Remember, $\varepsilon_{0}$, $\varepsilon_{r}$, $2\pi$ and $Q$ are real-valued constants.

So we can rewrite the integral:

$$V=\int_{a}^{b}{\frac{Q}{2\pi r\varepsilon_{0}\varepsilon_{r}}\:dr}=\frac{Q}{2\pi\varepsilon_{0}\varepsilon_{r}}\int_{a}^{b}{\frac{1}{r}\:dr}$$

We also know that $\int{\frac{1}{r}\:dr}=\ln{|r|}+c_{1}$, so we have:


share|improve this answer
Given that we can presume $r$ represents a (non-negative) radius, the modulus can be leaved apart. –  enzotib Aug 2 '12 at 17:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.