4
$\begingroup$

If there is a positive charge $q$ at the origin of a coordinate system, the electric potential $\phi$ at a distance $r$ from $q$ is (by definition, if we take the point of zero potential at infinity):

$$\phi=-\int_{\infty}^r \vec{E}\cdot d\vec{r}$$

The dot product of $\vec{E}$ and $d\vec{r}$ is $-E\text{ }dr$ because they point in opposite directions, so $$\phi=\int_{\infty}^r E\text{ } dr$$ For a positive point charge $q$, we have that: $$\phi=\frac{q}{4\pi\epsilon_0}\int_{\infty}^r \frac{dr}{r^2}$$ And evaluating the integral we arrive at: $$\phi=-\frac{q}{4\pi\epsilon_0 r}$$

However, the result should be positive according to Halliday-Resnick (fifth edition, page 608). They have the same derivation essentially, except that after evaluating the integral for some reason they get a positive potential. What's up?

$\endgroup$
5
$\begingroup$

The integrand $\vec E \cdot d\vec r$ is $E\,dr$, not $-E\,dr$.

The evaluation of the dot product is sort of done for you when you specify the curve on which you are integrating (i.e., your limits of integration in this case). You've double-accounted for the relative directions of $\vec E$ and $d\vec r$.

I suspect the underlying confusion is that you are treating $d\vec r$ as a small displacement along your path of integration. This is not correct. As you can see in this WP page on spherical coordinates, $d \vec r$ has a definition which doesn't depend on your integration path or direction: $d\vec r = dr\,\hat r + r\, d\theta\, \hat \theta + r\, \sin\theta\, d\phi\, \hat\phi$. Notice the unit vector $\hat r$, which always always points outward, away from your origin.

Here's the too-much-detail way to evaluate your integrand: $$\vec E(r) \cdot d\vec r = kq/r^2\,\hat r \cdot \left(d\vec r = dr\,\hat r + r\, d\theta\, \hat \theta + r\, \sin\theta\, d\phi\, \hat\phi\right)= kq/r^2\, dr = E(r)\, dr$$

$\endgroup$
  • 1
    $\begingroup$ People have this confusion about gravitational potential energy as well. It is a common misconception. $\endgroup$ – Gödel Aug 12 '14 at 1:48
  • 1
    $\begingroup$ Thanks so much, stupid of me. It's slightly upsetting that Halliday-Resnick did this incorrectly and then literally just changed the sign after integrating, but what're you gonna do... $\endgroup$ – Physics Llama Aug 12 '14 at 2:06
  • $\begingroup$ Hi Llama, I am facing the same problem. I have gone through the solution mentioned by BMS, but still don't get it. The potential at a point is always the work done per unit charge by an external force in bringing that charge from infinity to that particular point against the electrostatic force of repulsion. Since, the external force that does the work in this case is along the displacement, which makes the work done by the external force positive, the potential becomes negative after integrating this work from infinity to the point 'r'. Am I wrong? $\endgroup$ – Swami Sep 16 '14 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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