The relation between electric field and potential is given by: $$\vec{E}=-\triangledown{\phi}$$

Now, if we reverse the direction of $\vec{r}$, will the relation between electric field and potential change as: $$\vec{E}=\triangledown{\phi}$$

I mean will this equation ($\vec{E}=-\triangledown{\phi}$) also change if we change the $r$-coordinate system? Why or why not?


I think the equation would change because:

(A)\begin{align} \vec{E}(r)=-\triangledown{\phi} \tag{1} \end{align}

Also: \begin{align} k\dfrac{q}{r^{2}}(\hat{r})=-k\dfrac{q}{r^{2}}(-\hat{r})\\ \vec{E}(r)=-\vec{E}(-r) \tag{2} \end{align}

By (1) and (2):

\begin{align} -\triangledown{\phi}=-\vec{E}(-r)\\ \vec{E}(-r)=\triangledown{\phi} \end{align} (as the potential is a scalar and will not change if we reverse the $r$-coordinate system. $-$ see this link)

Am I wrong somewhere? Please explain a bit elaborately why this is not so.

  • $\begingroup$ I don't think your question makes sense if you don't specify the points at which you evaluate the terms in the equation (at $\mathbf{r}$ or at $-\mathbf{r}$?). Generally fundamental equations in physics like this should be independent of the choice of coordinate system (if that's what you mean by reversing the direction of $\mathbf{r}$). $\endgroup$ Commented May 8, 2017 at 9:46
  • $\begingroup$ Yes. That is what I mean. Will this equation ($\vec{E}=-\triangledown{\phi}$) also change if we change the $r$-coordinate system?............Why or why not? $\endgroup$
    – user143678
    Commented May 8, 2017 at 11:08
  • 1
    $\begingroup$ Why would the equation change? $\endgroup$ Commented May 8, 2017 at 13:55
  • $\begingroup$ @user1583209: I have edited the post. Please point out where am I wrong. $\endgroup$
    – user143678
    Commented May 9, 2017 at 5:35
  • $\begingroup$ I tried, see my answer below. $\endgroup$ Commented May 9, 2017 at 9:36

2 Answers 2


First let's see where the equation $\mathbf{E}=-\nabla \phi$ comes from. In a nutshell: you start from the electric field $\mathbf{E}$ which is a modification of space such that charges $q$ experience a force $q\mathbf{E}$. If you move this charge around the field along a path $C$ you change its potential energy by $$V=-q\int_C \mathbf{E}\cdot\mathbf{dl}$$ The quantity $$\phi (\mathbf{r})\equiv \frac{V}{q}=-\int_C \mathbf{E}\cdot\mathbf{dl}$$ is the electric potential and the path $C$ goes from a point with zero potential to $\mathbf{r}$. In the special case of conservative electric fields ($\nabla\times\vec{E}=0$, static fields, the integral above does not depend on the path $C$) you can write the electric field as $$\mathbf{E}(\mathbf{r})=-\nabla\phi(\mathbf{r})$$

Note that in this derivation nothing has been said about the chosen coordinate system or any directions. That's why this equation is general, independent of the choice of coordinate system. This is true for all fundamental laws in physics. Equations which depend on the chosen coordinate system you would only find if you study a concrete system which is asymmetric. However in this case these would be equations which apply to this concrete system only and not general equations.

Example point charge at origin

For a charge $q$ at the origin, the electric field is given by Coulomb's law as: $$\mathbf{E}(\mathbf{r})=\frac{kq}{r^2}\hat{\mathbf{r}}$$ where $\hat{\mathbf{r}}=\frac{\mathbf{r}}{|\mathbf{r}|}$ is the unit vector from the origin to the point $\mathbf{r}$. The corresponding electric potential is

$$\phi(\mathbf{r})= \frac{kq}{r}$$

and only depends on the distance $r$ from the origin (not on the direction), because the system is spherically symmetric.

The electric field satisfies the symmetry: $$\mathbf{E}(\mathbf{-r})=-\mathbf{E}(\mathbf{r}),$$ telling you that the electric field points in opposite directions and is of equal strength for points $\mathbf{r}$ that are point-symmetric to the origin.

The electric potential satisfies the symmetry: $$\phi(-\mathbf{r})=\phi(\mathbf{r})$$

In this spherical symmetric system, the gradient simplifies to:

$$\mathbf{E}(\mathbf{r})=-\nabla\phi(\mathbf{r})=-\frac{\partial\phi}{\partial r}\hat{\mathbf{r}}$$, which as you can see is satisfied by the expressions for $\mathbf{E},\phi$ above for all $\mathbf{r}$.

Change of coordinate system

If you transform to a new coordinate system where $$\mathbf{r}'=-\mathbf{r}$$ and consequently (handwaving physicist way of doing things, if you are a mathematician take a deep breath now... or look up Jacobian which would also work for more general coordinate transformations) $$\partial r' = - \partial r$$

you get

$$\mathbf{E}(\mathbf{r}')=\mathbf{E}(-\mathbf{r}) =\nabla\phi(\mathbf{r}) =-\nabla'\phi(\mathbf{r}')$$

where I have used $\nabla'=-\nabla$ to denote the gradient in the new coordinate system. So I guess your problem might be because you forgot to transform the gradient to the new coordinate system, which you should. As I suggested above, for general coordinate transforms the Jacobian will enter and you could try for yourself to see whether you can for instance rotate your coordinate system or whether you can transform to cartesian coordinates and still recover the same general equation $\mathbf{E}=-\nabla\phi$ (you should!).

(B)Also if you move in the direction of (−r) electric field and potential will increase.

I don't understand this argument. Could you elaborate what you mean?

  • $\begingroup$ (B) was a misconception. (I thought $\vec{E}=-\triangledown\phi $ means that when we go in the direction of $\vec{r}$, electric field and potential decreases.) Now I know that statement is wrong. I read here it instead means $-$in the direction of electric field, potential decreases. I see this holds for both positive and negative charges (i.e. moving in the direction of electric field, potential decreases.) $\endgroup$
    – user143678
    Commented May 9, 2017 at 11:30

We have, $$\textbf{E} (\textbf{r})= - \nabla\phi(\textbf{r}) \Rightarrow \textbf{E} (-\textbf{r})= - \nabla\phi(-\textbf{r}).$$

If $\phi(-\textbf{r}) = \phi(\textbf{r})$, then $\textbf{E} (-\textbf{r})= - \nabla\phi(\textbf{r}).$

If $\phi(-\textbf{r}) = -\phi(\textbf{r})$, then $\textbf{E} (-\textbf{r})= \nabla\phi(\textbf{r}).$

  • $\begingroup$ How can $\phi (-r)$ be $-\phi (r)$. Isn't $\phi$ a scalar and independent of the orientation of $\vec{r}$ $\endgroup$
    – user143678
    Commented May 8, 2017 at 8:15
  • $\begingroup$ If $\phi$ is an odd scalar function, then $\phi(-\vec{r}) = - \phi(\vec{r})$. $\endgroup$
    – rainman
    Commented May 8, 2017 at 8:28
  • 1
    $\begingroup$ Is electric potential really an odd scalar function? $\endgroup$
    – user143678
    Commented May 8, 2017 at 8:31
  • $\begingroup$ There is no such restriction. It can be any function. $\endgroup$
    – rainman
    Commented May 8, 2017 at 8:55

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