# The electric potential and the electric field

I'm preparing for my school exam in physics and I found that I don't understand the derivation of $E = -\nablaφ$. Here is a derivation from my book:

1) Imagine that 1 and 2 are infinitely close points with coordinates $x_1$ and $x_2$. They are so close that the electric field is a constant.

2) $dx = x_2 - x_1$

3) $-dφ = φ_1 - φ_2$

4) $Edx = -dφ \Rightarrow E = -\frac{dφ}{dx}$. Then we can generalize it on other coordinates (y and z) and be happy.

But I can't understand why $dφ$ does not equal $0$. It decreases more slowly than the electric field. So if the electric field is constant, the electric potential must be constant too.

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Electric field is like force while potential is like energy. Electric potential difference between two points is the amount of work done by electric field in taking a particle of unit charge from one point to the other i.e. $Edx=-d\phi$ (use of minus sign is by convention). Constant electric field doesn't imply constant potential. Compare it with the case when some object falls from a height h. During its fall gravity does work on it (thus increasing its speed), and the total work done (divided by the mass of object) is the potential difference between the ground and the initial position. – user10001 May 2 '13 at 18:05
So even though gravitational field can be taken almost constant near the surface of earth, the potential varies. The situation in case of electric field is same. Here role of gravitational field is played by electric field, and role of mass is played by electric charge. – user10001 May 2 '13 at 18:05

Since you are a high school student I'm going to make this as simple as possible. This should be read in addition to joshphysics's (correct) answer. The purpose of my own answer is to specifically fill in the steps that the book left out, but which are nevertheless correct.

First, I will take a guess that you have seen basic integral calculus and that your book has already defined the electric potential (in one dimension, for example $x$) in the following way: $$\int_{x_1}^{x_2}E(x)dx=-\Phi(x_2)+\Phi(x_1).$$ For any well-behaved function $$\lim_{\Delta a \rightarrow 0}\int_{a}^{a+\Delta a}f(x)dx=\lim_{\Delta a \rightarrow 0}f(a)\Delta a,$$ since we are evaluating the area of one thin rectangle with height $f(a)$ and width $\Delta a$. Anyway, this leads to, when $x_2$ is very close to $x_1$ (so that $x_2 = x_1 + \Delta x$ with $\Delta x$ becoming small): $$\lim_{\Delta x \rightarrow 0}\int_{x_1}^{x_1+\Delta x}E(x)dx=\lim_{\Delta x \rightarrow 0}E(x_1)\Delta x=\lim_{\Delta x \rightarrow 0}\left(-\Phi(x_1+\Delta x)+\Phi(x_1)\right).$$ Continuing with this last step, we find $$E(x_1)=\lim_{\Delta x \rightarrow 0}\left(-\frac{\Phi(x_1+\Delta x)-\Phi(x_1)}{\Delta x}\right)$$ or, since $x_1$ was arbitrary, $$E(x)=\lim_{\Delta x \rightarrow 0}\left(-\frac{\Phi(x+\Delta x)-\Phi(x)}{\Delta x}\right)=-\frac{d\Phi(x)}{dx}.$$ The book is right that this can be extended to the variables $y$ and $z$.

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Thank you very much. If I could, I would rate your answer positively, but I can't :( . – cheremushkin May 2 '13 at 19:04
Cool. If you want to accept it as the answer, just click the check-mark. – Pricklebush Tickletush May 2 '13 at 19:05

As far as I can tell, this "argument" is not a meaningful derivation of $\mathbf E = -\nabla \Phi$. In step 4, one seems to assume that $Edx = -d\phi$ which is essentially what one is trying to prove in the first place, so the argument seems circular.

The precise derivation of this fact relies on the crucial fact that the electric field is a conservative vector field which then immediately guarantees the existence of a scalar function $\Phi$ for which $\mathbf E = -\nabla\Phi$.

Notice that if the electric field is constant in some region then the gradient of $\Phi$ is a constant there. In particular its partial derivatives satisfy $$\partial_x\Phi = -E_x, \qquad \partial_y\Phi = -E_y, \qquad \partial_z\Phi = -E_z$$ The $x$ equation gives $$\Phi(x,y,z) = -E_x x + f(y,z)$$ Then the $y$ equation gives $\partial_yf = -E_y$ which implies $f(y,z) = -E_y y + g(z)$ and finally the $z$ equation gives $\partial_z g = -E_z$ so that $g(z) = -E_z z + c$. Putting this all together gives $$\Phi(x,y,z) = -\mathbf E \cdot \mathbf x + c$$ In particular, the electric field being constant does not imply that the electric potential is constant, it means that it has the general form written in this last equation.