3
$\begingroup$

I thought the relation between the electrostatic field $\vec E$ and the electrostatic potential $V$ is as follows:

$$\vec E = - \nabla V$$

Thus, when $V$ is zero, $\vec E$ is also zero.

$\endgroup$
  • 1
    $\begingroup$ Do you mean $V$ is zero identically or that there's a value of $\vec{x}$, so that $V=0$? $\endgroup$ – auxsvr May 10 '15 at 22:15
5
$\begingroup$

It depends on what you mean when you say $V=0$. In the context of the equation:

$$\vec{E}=-\nabla V$$

which holds specifically in electrostatics $V$ is a scalar field, meaning that it is actually a function which assigns every point in space a scalar value. $\vec{E}$ is a vector field, which assigns a vector to every point in space. Thus, both the electric field and the potential are dependent upon position. This can be shown more explicitly as:

$$\vec{E}(\textbf{r})=-\nabla V(\textbf{r})$$

where $\textbf{r}$ is a position vector. Now, if $V(\textbf{r})=0$ for all $\textbf{r}$ then certainly the gradient is also zero everywhere, and thus, the electric field is zero everywhere.

On the other hand $V(\textbf{r})$ may equal zero for only some $\textbf{r}$. For example, at the point $P$ midway between two point charges, one with charge $+q$ and the other with charge $-q$ the potential is zero, assuming infinity as the reference point. However, if you move even slightly away from this point, the potential is non zero. The fact that the potential is changing at point $P$ indicates that the gradient at this point is non zero. Thus, the electric field at $P$ is non zero, even though the potential itself is zero at $P$.

Note that this is true for electrostatics, but, as Sebastian mentions in a comment below, it is incomplete in the context of electrodynamics. This is simply because the expression you point to relating the electric field to the potential only holds for electrostatics. For treatment of the more general case, please see Sebastian's and Alexander's fine answers.

$\endgroup$
  • $\begingroup$ This is an incomplete explanation. Although it is more didactic an clear about one point. As the question does not specify it relates to electro-statics, one has to mention that the formula $E = -\nabla \phi$ is not the whole truth (compare my and Alexander's answers)! $\endgroup$ – Sebastian Riese May 11 '15 at 8:10
  • $\begingroup$ @Sebastian Riese You are right, I should have been more clear to this end. Thank you for pointing that out. I have edited my answer to be more clear, but have chosen to point to yours and Alexander's answers for the case of electrodynamics rather than include it myself, as I do assume the OP is considering electrostatics, given that the expression he/she is using is true only in this case. Besides, why reinvent the wheel? There are already two fine answers addressing this point ;) $\endgroup$ – wgrenard May 11 '15 at 11:37
1
$\begingroup$

This is up to a gauge you're using.
The most general expression is -
$$ \vec{E}=-\nabla\phi-\frac{\partial\vec{A}}{\partial t} $$ $$ \vec{B}=\nabla \times \vec{A} $$

so it's possible to nullify $\phi $ and still get any electromagnetic field you want.

$\endgroup$
1
$\begingroup$

The relationship between electrical field and scalar potential you give only holds in the static case (or after an gauge transformation).

The full formula for the case of time dependent fields is: $$ \vec E = - \nabla V - \partial_t \vec A $$ The magnetic field then is given by: $$ \vec B = \nabla \times \vec A. $$

$\endgroup$
0
$\begingroup$

$$\vec{E}=-\vec{\nabla}V$$ means: the electric field is the derivative (3 dimensional) of V. So wether $V = 0$ or not, doesn't matter. It's wether V changes or not, that defines the electric field.

An analogy to make things clear: analogy for potential = height, analogy for electric field= 'how does the ball roll?'. To see 'How does the ball roll?', you don't need to know the absolute height, only wether there is a slope (a difference in height): an equal slope has the same effect on the ball in the Netherlands as in Nepal.

$\endgroup$
  • $\begingroup$ I assume the question is about $\forall_x V(x) = 0$, not about $V(x_0) = 0$ $\endgroup$ – Sebastian Riese May 10 '15 at 22:51
0
$\begingroup$

Electric field is vector so there is a possibility for the electric field to be zero at a point but it isn't the same with the electric potential it is a scalar ie the net potential is the algebraic sum of individual potentials so it is not necessary for potential to be zero if field is zero and vice versa hope u understood

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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