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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

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  • $\begingroup$ 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. $\endgroup$
    – Adam Sturge
    Mar 18 at 3:16

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

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  • $\begingroup$ 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 $\endgroup$
    – Adam Sturge
    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 $\endgroup$
    – Adam Sturge
    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 $\endgroup$
    – Adam Sturge
    Mar 19 at 16:29

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