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When a line of charge has a charge density $\lambda$, we know that the electric field points perpendicular to the vector pointing along the line of charge.

When calculating the difference in electric potential due with the following equations.

$\nabla V=-\vec{E}$

Therefore

$\Delta V = -\int_{\vec{r_o}}^\vec{r_f}E\cdot \vec{dr}$

knowing that

$\vec{E} = \frac{\lambda}{2\pi\epsilon_or}\hat{r}$

and that

$\left\lVert\vec{r_f}\right\lVert < \left\lVert\vec{r_o}\right\lVert $

Carrying out the integration (Hopefully correctly) I got.

$\Delta V = \frac{\lambda}{2\pi \epsilon_o} ln(\frac{r_f}{r_o})$

What confuses me is that the ln() is negative. I assume that the value should be positive since we move closer towards the line of charge should give us a positive change in electric potential. My best guess for my problem is that I missed a negative somewhere, but looking at online solutions they've got the same answer that I got.

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    $\begingroup$ You missed the minus sign. The direction of changing you position is taken care of purely by the limits of integration, NOT by any sign on d$\vec{r}$. $\endgroup$ – Bill N Mar 5 '18 at 20:07
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No, it's okay. The pontential difference increases as you go farther. The less you move away, the more similar potential you have (little difference).

By the way

  1. You can't integrate in three dimensions that way. You're using cylindrical coordinates (because of the symmetry of the problem), and you integrate along $r$, which is $|\vec{r}|$.
  2. The limits of integration are thus scalars. However, $\vec{E}$ is a vector, and you do the scalar product inside the integral, but fortunately the angle is 0 degrees.
  3. You missed the minus sign in front of the integral, so it appears outside the $\ln$. Was that your question? Because now

$$\Delta V = -\dfrac{\lambda}{2\pi\varepsilon_0} \ln \left(\frac{r_F}{r_o}\right)$$

and the voltage difference increases when you go further, but in a negative sense, which means it becomes "more negative" as you move away.

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To elaborate a bit on Bill's comment, you might consider a curve defined as follows in some cylindrical $(r,\theta,z)$ coordinate system:

$$\gamma(t) = \big(r(t),\theta(t),\phi(t)\big) = (t, 0, 0)$$ $$ t \in [r_0,r_f]$$

The tangent vector to this curve is

$$\frac{d\vec r}{dt} = \hat r $$

so

$$\Delta V = -\int_\gamma \vec E \cdot d\vec r = -\int_{r_0}^{r_f} \vec E \cdot \frac{d\vec r}{dt} dt = -\frac{\lambda}{2\pi\epsilon_0}\int_{r_0}^{r_f} \frac{dt}{t} = -\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r_f}{r_0}\right) $$

$$ =\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r_0}{r_f}\right) $$


Whenever things like this happen, I find it useful to introduce an explicit, unambiguous parameterization of my curve, which usually resolves the issue.

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  • $\begingroup$ Am I missing something, or you have an extra $dt$ on your 4th equality? ($\dfrac{dt}{t} dt$ shouldnt this just be $\dfrac{dt}{t}$?) $\endgroup$ – Aram May 23 at 18:38
  • $\begingroup$ @Aram Yes, you are right. $\endgroup$ – J. Murray May 24 at 0:38

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