I know Work= minus Electric Potential Energy. So you have: $-q\ dV=-qE\ dx$ and thus $E= -dV/dx$.

But does that 'minus' sign have any effect so $\mathbf E$ vector's direction.

Say we have the fields lines (red), equipotential lines (blue), between two charged plates (black). So $E$ is $3-0=-3$ V/m. What does that minus supposed to tell me?

enter image description here

  • 1
    $\begingroup$ Actually given the electric field is a vector it's more like $\vec E=-\nabla V$. $\endgroup$ Commented Sep 12, 2017 at 23:56

5 Answers 5


The minus sign sure does have an effect on the electric field's direction! It's set up that way so that when you compute your E-field, it will correctly follow the convention: E-field points "downhill" relative to the potential. That is, it comes from positive charge, and goes to negative charge.

Additional explanation added in edit:

So for your example, the voltage slopes down from the positive plate to the negative plate. If you ascribe $-dV/dx$ to the $x$-component of the electric field, the field will point "downhill" because it will be negative when the slope is positive. Thus, the field points from the positive plate to the negative plate. But keep in mind that this decision to have E slope "downhill" is arbitrary. The convention could have been the other way.

  • $\begingroup$ The difference in potential is always the voltage of the positive plate minus the voltage of the negative plate. But how do you know that minus sign tells you it goes from positive to negative. I mean I know it is so because my teacher tells me. Not because I look at the sign andthe math and be like aha! it is from positive to negative. $\endgroup$ Commented Sep 13, 2017 at 1:21
  • $\begingroup$ @user132522 I added a little more to the answer. Does this help? $\endgroup$
    – Gilbert
    Commented Sep 13, 2017 at 3:27
  • $\begingroup$ how to compute dV? Do you substract the voltage of a point closer to the negative plate, from the voltage of a point farther from the negative plate, so the result is always positive (I am assuming dx will always be positive)? $\endgroup$ Commented Sep 14, 2017 at 21:43

In fact, $\mathbf{E}$ is not just $-\frac{dV}{dx}$. There is a little bit more than that.

A Little bit of math

In mathematics, we have the concept of a vector field: a region of space where, in every point, there is a vector pointing somewhere. This can be represented as a function, for example: $\vec{f}(x,y) = \begin{bmatrix} x \\ y \end{bmatrix}$. This would mean that, for example, in $(1,0)$ the vector is $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. And you can imagine in for all points in the plane. This is what happens in space when you have, for example, an ponctual charge.

Now it is important to talk about the meaning of potential. We know that we have potential in many areas of physics, from mechanics to electromagnetism. And indeed, these potentials are very, very related to vector fields! In fact, we know that a potential is a scalar, so it can be represented as a scalar function, for example $\phi(x,y) = 2x+y$. Now, how do we relate some scalar potentials and vector fields? We extend the notion of derivative using the gradient. The gradient of a scalar function $\phi(x,y,z)$ is defined as follows (in 3D space): $$\nabla \phi = \frac{\partial \phi}{\partial x}\mathbf{î} + \frac{\partial \phi}{\partial y}\mathbf{\hat{j}} + \frac{\partial \phi}{\partial z}\mathbf{\hat{k}}$$ And, as you can see, the gradient is a vector. More than that, the gradient is a vector field. And notice we are using derivatives here: This vector fields points to the direction of where the potential changes the fastest. Now that we have the basic mathematical background, let's talk about physics.

And Physics

We know, now, that we can create vector fields from potentials. And this is done with the use of gradients. Now, suppose you have a one dimensional case: the surface of earth. You can define, here, the potential $V(z)$ as function of your altittude $z$. This is known, and we define it as $$V(z) = mgz$$ Which is the potential energy. Now, suppose we would define the vector field of weight forces as $\mathbf{F} = \nabla V$. This would mean that "the vector $\mathbf{F}$ must point to the direction of increasing potential". This conclusion comes directly from our definition of gradient. And that would mean that waterfalls would actually go upwards! More than that, everything would go upwards to where is the "greater potential" region. So, we simply define that $$\mathbf{F} = - \nabla V$$ And this must also be done with a charge in space when there is electric potential. If we had defined that $\mathbf{E} = + \nabla V$, then we would have positive charges going from regions with fewer potential to more potential. This would mean that equal charges would attract each other and different ones would repel each other. It would also make lightning bolts go upwards (from earth to sky) instead of from sky to earth. That is why we define $$\mathbf{E} = - \nabla V$$ or, in the 1-dimensional case, $$\mathbf{E} = - \frac{dV}{dx}$$


Okay, So I'm putting it in the simplest way I can

Referemce We all know that Electric Field Lines start from the positive charges and Terminate at Negative charges. So E = - dV/dr simply implies that as you move in the direction of electric field, electric potential decreases. And it is quite obvious because as you are going in the direction of EF, you are in a way moving towards negative charges so electric potential is bound to decrease.


It might help you to think that the minus sign means stuff "rolls down hill". The gravitational force is positive in the downward direction, towards decreasing (more negative) gravitational energy. It's analogous with electricity: positive electric field toward more negative electric potential.


The minus comes from the definition of the electric potential at a point and to simplify matters that there is only motion along the x-axis.

There are two variations:

  1. The electric potential at a point is the work done by an external force acting on a unit positive charge in taking the charge from a position where the potential is zero to the point.
  2. The electric potential at a point is minus the work done by the electric field acting on a unit positive charge in taking the charge from a position where the potential is zero to the point.

Consider a unit positive charge in a uniform electric field $E$ where the electric field is in the positive x-direction.
There is a force $E$ acting on the positive charge in the direction of the electric field.

Allowing the positive charge to move by a displacement $dx$ in the postive x-direction.
The electric field does work on the charge $E\,dx$.

By definition 1, minus this amount of work done by the electric field is equal to the change in electric potential of the charge $dV$.

So you get $-E\,dx = dV$ which leads to the equation $E = - \dfrac {dV}{dx}$

Now look at it from the point of view of an external force which is acting on the unit positive charge.
That force must be acting in the negative x-direction and have a magnitude $E$ so that the net force of the charge is zero (ie the charge does not accelerate).
In moving the positiver charge a distance $dx$ in the positive x-direction the work done by the external force is $- E \, dx$.
The negative sign comes from the fact that the force and the displacement are in opposite directions so the dot product of force and displacement (= work done) is negative.
This again changes the electric potential of the charge by $dV$

So again you get $-E\,dx = dV$ which leads to the equation $E = - \dfrac {dV}{dx}$

Your diagram illustrates these ideas very well.

enter image description here

The electric field is downwards and so it is positive.

The potential gradient $\dfrac{dV}{dx} = \dfrac{V_{\rm final}-V_{\rm initial}}{dx}$ is negative because as you are going down ans $V_{\rm final}<V_{\rm initial}$.

This means that $-\dfrac{dV}{dx}$ is positive and so therefore is E positive ie in the direction of increasing $x$.


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