0
$\begingroup$

I've been trying to figure this out on my own for days and finally decided to start googling. For reference I'm self studying Griffith's E&M textbook. I'm at the part where he derives the equation for the electric potential V of a point charge at the origin. Which he gives as $\frac{1}{4\pi\epsilon}\frac{q}{r}$. He then goes on to say that you can use the superposition principle to move from this to the sum $\frac{1}{4\pi\epsilon}\sum_{i=0}^{n}\frac{q_{i}}{r}$ Where here $r$ is understood to be the distance from the source charge to the test point

However this forumula, blindly applied, gives rise to a commonly asked question on this and other forums. If you use it to compute the potential at the midpoint between 2 opposite charges you get 0. And this together with $\vec{E} = -\nabla V$ gives a 0 for the electric field. Clearly this is wrong since the particle should be moving away from the positive charge towards the negative one.

Since we know the Electric field of a point charge we can compute the net field directly via superposition. Since the negative charge has a separation vector that is opposite that of the positive charge the expected answer is $\vec{E} = \frac{1}{\pi\epsilon}\frac{2q}{d^{2}}\hat{d}$. Where $d$ is the separation distance between the 2 charges (making our test point at $\frac{d}{2}$)

So what gives? How is the answer simultaneously 0 and not 0? Other posts on this topic have skirted around answering the exact question I have. Which is all these equations are held to be universally true. I understand that the electric field is not actually zero. I can visualize the inflection point between the hill of positive potential and the valley of negative potential. But just from raw analytic calculation I would expect these equations to agree

$\endgroup$
3
  • $\begingroup$ If you use it to compute the potential at the midpoint between 2 opposite charges you get 0. That doesn’t mean the field at the midpoint is zero. You can’t compute the gradient from a value at one point. $\endgroup$
    – Ghoster
    Mar 18 at 3:05
  • $\begingroup$ "How does 0 electric potential but non-zero electric field work mathematically?" Roughly analogous to how you can have 0 velocity but non-zero acceleration. In general, the value of a function at a point does not determine the value of that function's derivative at that point. $\endgroup$
    – hft
    Mar 18 at 3:08
  • $\begingroup$ I think I'm beginning to understand. The equation I used from Grifiths gives me the value at the midpoint. But to compute the Electric field from the potential via the gradient I need the full potential equation for all space. $\endgroup$ Mar 18 at 3:16

1 Answer 1

3
$\begingroup$

The value of the potential doesn't matter, only its gradient.

$\endgroup$
4
  • $\begingroup$ The value I got is 0. It could be shifted by a constant sure depending on where I chose my reference point. But the gradient of a constant is still 0. My confusion is over the differing values for E not the value I got for V $\endgroup$ Mar 18 at 3:13
  • $\begingroup$ I think I'm beginning to understand. The equation I used from Grifiths gives me the value at the midpoint. But to compute the Electric field from the potential via the gradient I need the full potential equation for all space $\endgroup$ Mar 18 at 3:16
  • $\begingroup$ @AdamSturge, have you ever been to the Dead Sea or Death Valley? At some point you stood on (or drove over) a point with zero elevation (according to our conventional reference of sea level), but with a nonzero slope (otherwise you couldn't have descended further). A position with 0 potential but nonzero field is analogous. $\endgroup$
    – The Photon
    Mar 18 at 15:33
  • $\begingroup$ The concept of 0 potential and non zero field made sense to me physically. As I said I can picture the surface with a hill on one side and a valley on the other. I was confused about the mathematics. Why two equations that purported to equal the same thing gave different answers. As was pointed out above I was differentiating a point on the potential surface not the whole surface. A silly mistake in hindsight $\endgroup$ Mar 19 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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