# How does 0 electric potential but non-zero electric field work mathematically

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

• 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. Commented Mar 18, 2023 at 3:05
• "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.
– hft
Commented Mar 18, 2023 at 3:08
• 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. Commented Mar 18, 2023 at 3:16

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

• 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 Commented Mar 18, 2023 at 3:13
• 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 Commented Mar 18, 2023 at 3:16
• @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. Commented Mar 18, 2023 at 15:33
• 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 Commented Mar 19, 2023 at 16:29