How does $E = -\nabla V$ work for a dipole? [closed]

For a dipole, the potential is zero at any point on the perpendicular axis, but the electric field isn't zero along that axis. Now, how do I get $E$ from $V=0$ (at that point/axis) from the relation:

$$E = -\nabla V.$$

• Welcome on Physics SE :) See physics.stackexchange.com/help/notation for formatting formula. Also, at least I am struggling to understand what exactly your claim is :/ do you have a source for it? – Sanya Aug 16 '16 at 7:20
• Hello Abhishek. I have added a bit of math formatting to your question. Also, see Sanya's link. Please let me know if something isn't as intended – Steeven Aug 16 '16 at 7:35
• It's not really clear to me what the question here is, or why it got two upvotes. You get $E$ from that relation by computing $\nabla V$, obviously. The derivative of a function at a point where it is zero can be non-zero, this has nothing to with physics. – ACuriousMind Aug 16 '16 at 10:28
• well am sorry i am not that good . As per to why it got upvotes?..i dont know how it works either. – abhishek bhat Aug 16 '16 at 11:20
• In principle I agree with @ACuriousMind, on the other hand the problem of separating math and physics affects plenty of posts here, including several of his own top answers. Eventually, it boils down to a symmetry problem here. – mikuszefski Aug 16 '16 at 11:45

While I agree with the other answers (confusing value of a function and the gradient as stated by @jayjay as well as the helpful image by @john-rennie), I am missing the dipole and/or the axis description, though. The Potential of an (electric) dipole with stength and direction $\vec P$ is given by:

$$V(\vec r)=-\frac{1}{4\pi \varepsilon_0}\vec P \cdot (\vec\nabla\frac{1}{r})=\frac{1}{4\pi \varepsilon_0}\frac{\vec P\cdot \vec r}{r^3}$$

So you can easily see that $V=0, \forall \vec r \perp \vec P$ (apart from $r=0$ of course).

The field is, as mentioned in the question:

$$\vec E(\vec r)=-\nabla V(\vec r)=-\frac{1}{4\pi \varepsilon_0} (\frac{\vec P}{r^3}-3\frac{(\vec P\cdot \vec r)\vec r}{r^5})$$

The part with the factor 3 again vanishes if $\vec r \perp \vec P$, but $\vec P/r^3$ remains giving a field opposite to $\vec P$. Hence, the fact that $V=0$ in the plane perpendicular to $\vec P$ reflects in the fact that there is no field component within the plane. That would be the "constant potential means no field" claim, but only for 2 out of 3 dimensions.

You are making the mistake of thinking of the gradient as a regular one dimensional function where you pop in a value and it throws out an output. You can't take the gradient of a number (in your case, $0$). You take the gradient of a function (which can have as many dimensions as you like -- a multivariable function, that is). Now, the potential function of a dipole is

$$V = \frac{p \cos(\theta)} {4\pi\varepsilon_0r^2}$$

Take the gradient of this function (in spherical coordinates) and you'll get the electric field as a function of $\theta$ and $r$. Now pop in $\theta=\pi/2$ and you'll get $V=0$ and the electric field as a function of $r$ (everywhere on the perpendicular axis).

• Hello JayJay. I added math formatting to the answer - please check that everything still is as intended. – Steeven Aug 16 '16 at 7:28
• Gee thanks Steeven! Yes, everything is as intended – GeeJay Aug 16 '16 at 7:30
• i understand i cannot take the gradient of a constant, but at that point V is a constant, so i was wondering how to gt the scalar function(wch you stated in your answer). – abhishek bhat Aug 16 '16 at 8:06
• @abhishekbhat That's easy to find. You add the potential from each point charge, do some algebraic manipulation, add a pinch of approximation and voila! A quick Google search should throw up the derivation. – GeeJay Aug 16 '16 at 8:28

If you look at this diagram:

then it should be obvious that while $V$ is constant along the horizontal line it varies along the vertical line. That means the horizontal component of $\nabla V$ is zero and the vertical component is non-zero. So on the horizontal line $\mathbf E$ is a vector directed vertically upwards.

I assume that the dipole consists of two opposite charges separated a distance $d$ placed on the $z$-axis and centered at the origin. This gives a potential of \begin{align} V(x,y,z) = \frac{q}{4\pi\varepsilon_0 \sqrt{x^2+y^2+(z+\frac{d}{2})^2}} - \frac{q}{4\pi\varepsilon_0 \sqrt{x^2+y^2+(z-\frac{d}{2})^2}}~, \end{align} which will be zero if you are on the $xy$-plane (because $z=0$). Mathematically, it is \begin{align} V (x,y,0) = 0~. \end{align}

Now, when you want to find the electric field, you take the gradient of $V$, which is correct (i.e. $\vec{E}=-\nabla V$), but you would have to take the gradient of the general potential $V(x,y,z)$ and not the potential at the perpendicular (when $z=0$). Explicity, we want \begin{align} \vec{E}(x,y,z) &= -\nabla V(x,y,z) \\ &=\frac{q}{4\pi\varepsilon_0}\left(\frac{x \vec{e}_x + y \vec{e}_y + (z+\frac{d}{2})\vec{e}_z}{\left(x^2+y^2+(z+\frac{d}{2})^2\right)^{3/2}} - \frac{x \vec{e}_x + y \vec{e}_y + (z-\frac{d}{2})\vec{e}_z}{\left(x^2+y^2+(z-\frac{d}{2})^2\right)^{3/2}} \right) \end{align} and evaluating this at $z=0$ gives \begin{align} \vec{E}(x,y,0) =\frac{q}{4\pi\varepsilon_0} \frac{d \vec{e}_z}{\left(x^2+y^2+(\frac{d}{2})^2\right)^{3/2}} ~, \end{align} which is non-zero, as expected.

(Note: The calculation was done in Cartesian coordinates to avoid any assumptions about the dipole approximation.)