I learned that electric field strength is the negative of the potential gradient, i.e. $$E=-\frac{dV}{dr},$$ where $V$ is the potential difference and $r$ is the distance. The equivalent expression for gravitational field strength is $$g=-\frac{d\phi}{dr}$$ where $\phi$ is the gravitational potential.

Now, if there is an increasing electric potential with an increase in distance, the potential gradient $\frac{dV}{dr}$ will be positive, and hence $-\frac{dV}{dr}$ will be negative.

However, electric field strength cannot have a negative value as it is the force per unit positive charge acting on a test charge at a point in the electric field.

Therefore, I cannot resolve this seeming contradiction.

I tried to interpret the negative sign as an indication of direction instead of a representation of the magnitude of the electric field. In this case, the negative sign means the electric field is always in the opposite direction of the change of potential, which I am not sure is the correct interpretation.


3 Answers 3


Sign of the projection of a vector

It something meaningless to talk about the sign of a vector quantity, like the electric field $\mathbf{E}(\mathbf{r})$. You could talk about the sign of the projection of a vector along a direction indicated as an example by another unit vector, $\mathbf{\hat{t}}$.

A vector can be written as the product of a non-negative scalar (this is what is positive by definition, not the vector), namely its magnitude $|\mathbf{E}|$ and a unit vector pointing in the same direction of the vector $\mathbf{\hat{E}}$, usually indicated with a "hat",

$\mathbf{E} = |\mathbf{E}| \mathbf{\hat{E}}$.

If you choose another unit vector $\mathbf{\hat{t}}$, and project $\mathbf{E}$ along $\mathbf{\hat{t}}$, you get

$\mathbf{E} \cdot \mathbf{\hat{t}} = |\mathbf{E}| \mathbf{\hat{E}} \cdot \mathbf{\hat{t}} = |\mathbf{E}| \cos \theta$,

being $\theta$ the angle between the two directions indicated by the two unit vectors $\mathbf{\hat{E}}$ and $\mathbf{\hat{t}}$. Obviously, if you reverse the unit vector along which you perform the projection, putting a minus sign in front of it, this projection changes its sign as well,

$\mathbf{E} \cdot (-\mathbf{\hat{t}}) = |\mathbf{E}| \mathbf{\hat{E}} \cdot (-\mathbf{\hat{t}}) = |\mathbf{E}| \cos (\pi - \theta) = - |\mathbf{E}| \cos \theta$.

Work in electrostatics

Now, let's come back to the electric field in electrostatics, that can be written as the opposite of the gradient of a scalar potential field $V$,

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

and can be interpreted as the force per unit charge acting on a point charge, $\mathbf{F} = q \mathbf{E}(\mathbf{r})$, and let's compute the work done by the electric field on a charge along a path $\ell_{\mathbf{r}_0 \rightarrow \mathbf{r}_1}$ going from $\mathbf{r}_0$, to $\mathbf{r}_1$,

$W = \displaystyle \int_{{\ell}_{\mathbf{r}_0 \rightarrow \mathbf{r}_1}} \mathbf{F} \cdot d \mathbf{r} = q \displaystyle \int_{{\ell}_{\mathbf{r}_0 \rightarrow \mathbf{r}_1}} \mathbf{E}(\mathbf{r}) \cdot d \mathbf{r} = - q \displaystyle \int_{{\ell}_{\mathbf{r}_0 \rightarrow \mathbf{r}_1}} \nabla V(\mathbf{r}) \cdot d \mathbf{r} = - q \displaystyle \int_{{\ell}_{\mathbf{r}_0 \rightarrow \mathbf{r}_1}} d V(\mathbf{r}) = -q \left[ V(\mathbf{r}_1) - V(\mathbf{r}_0) \right]$.

This expression is the integral, i.e. the infinite sum of infinitesimal contributions, along the path $\ell_{\mathbf{r}_0 \rightarrow \mathbf{r}_1}$. Recalling what we said before about the sign of the projection of vectors, you can easily see that an elementary contribution to work done by the electric field (for a positive charge $q>0$), depends on the relative direction of the local electric field $\mathbf{E}(\mathbf{r})$ and the elementary displacement $d\mathbf{r}$, namely

$dW = q \mathbf{E}(\mathbf{r}) \cdot d \mathbf{r} = - q dV \left\{ \begin{matrix} > 0 \quad \text{if the projection of $\mathbf{E}(\mathbf{r})$ over $d\mathbf{r}$ is positive} \\ = 0 \quad \text{if these two vectors are orthogonal} \qquad \qquad \\ < 0 \quad \text{if the projection of $\mathbf{E}(\mathbf{r})$ over $d\mathbf{r}$ is negative} \end{matrix} \right.$

Convetions in the potential sign

There is no mathematical reason to prefer the definition of the potential as

$\mathbf{E} = - \nabla V$,

instead of

$\mathbf{E} = \nabla V$.

You would end up with a theory with opposite signs, wherever the potential appears with odd powers, since physics meant as nature fortunately is independent on our choice for the conventions.


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

$$\vec{E} =- \begin{bmatrix} \frac{\partial V}{\partial x}\\\frac{\partial V}{\partial y}\\\frac{\partial V}{\partial z} \end{bmatrix}$$

Force is a vector, a negative of this vector is in the opposite direction to the first.

Why makes you think that the force cannot be opposite direction to the gradient of potential?

Proof of this definition;

You can start many ways, however starting with the former , we can show it satisfies the definition of potential.

$$\vec{E} = - \nabla V$$ $$\int_{a}^{b}\vec{E} \cdot \vec{dr} = - \int_{a}^{b} \nabla V\cdot \vec{dr}$$

By definition of the gradient of a scalar function.

$$-\int_{a}^{b}\vec{E} \cdot \vec{dr} = V(b)-V(a)$$

Hence the former equation is the same as the definition of potential.

We have yet to prove that E is infact a Conservative field and thus can be written in this way, which can be done by either direct computation of the line integral and showing it is path independant. Or by taking the curl and showing it is zero

  • $\begingroup$ My question is why the force is in the opposite direction to the potential gradient; in other words, how do we interpret the negative sign in the equation 𝐸⃗ =−∇𝑉? $\endgroup$
    – Lu_xx
    Commented Dec 6, 2022 at 9:48
  • $\begingroup$ The electric field lines point in the direction of steepest decrease of potential, aka when work is done, potential energy decreases. This is energy conservation $\endgroup$ Commented Dec 6, 2022 at 10:15

Consider a point charge with charge $-|q|$ (some arbitrary negative value). Its electric field is given by $\vec{E} = -\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$. This is an electric field with negative field strength.

The negative sign in the electric field, to my understanding, encodes the fact that positive charges are repelled by positive charge distributions and that positive charges are attracted by negative charge distributions (and vice versa).

For example, using the electric field of a negative point charge, if we place a negative test charge $-|q_0|$ somewhere arbitrary, it will feel a force pointing in the $\hat{r}$ direction, i.e., radially outward and away from the negative charge.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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