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I know gradient shows you the direction in which function is increasing most. So does electric field direction show us in which direction potential energy function is decreasing most.... My efforts : we know electric field lines go from positive to negative charge. So that's where value of electric potential energy function is decreasing. As we know electric field lines show us the electric field direction. So electric field points out the direction in which potential energy function is decreasing so it's the negative gradient vector of the potential energy function. Am I right with this reasoning?

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Yes, you're right. Indeed we write$$\vec{E}=-\vec{\text{grad}}\ \phi\ \ \ \ \ \text{that is}\ \ \ \ \vec{E}=-\vec{\nabla}\ \phi$$

in which$$\vec{\nabla}\ \phi=\vec {i}\frac{\partial \phi}{\partial x}+\vec {j}\frac{\partial \phi}{\partial y}+\vec {k}\frac{\partial \phi}{\partial z}$$

This is the case because we know that the fall in potential is the work done per unit charge by the field, so$$d\phi=-\vec{E}.d \vec{r}\ \ \ \ \text{but also we know}\ \ \ \ d\phi= \frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz$$

which we can write as$$d \phi=\left(\vec {i}\frac{\partial \phi}{\partial x}+\vec {j}\frac{\partial \phi}{\partial y}+\vec {k}\frac{\partial \phi}{\partial z}\right).\left(dx\ \vec{i}+dy\ \vec{j}+dz\ \vec{y}\right)=\vec{\text{grad}}\ \phi.d\vec{r}$$

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I think the best way to see this is to read it the other way. The Electric Potential is a function such that the Gradient in a point is equal to the Electric Field in that point.

This is useful because then $\int_{\gamma (a)}^{\gamma (b)}E(\gamma (t))\cdot \gamma {(t)}'dt = V(\gamma (a))-V(\gamma (b))$.

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    $\begingroup$ You're missing the word "negative" somewhere. $\endgroup$
    – Bill N
    Commented Sep 30, 2018 at 2:32
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We know that the potential due to a positive charged particle at any point is given by $$V=\frac {q}{4π\epsilon r}$$

The potential is positive in case of positive charges because we have to do work to move a test charge from infinity to that point. Now if you calculate gradient of potential it will give the direction of maximum increase of potential but the potential will increase if we go towards the central charge because two positive charges repel each other therefore direction of force should be in the direction of the of potential. Therefore $E$ is given by -ve gradient of potential.

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I'd say the best way to describe it is by the definition of potential difference.

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

Because we know that the result is $V(b)-V(a)$ we know that the inside function is a conservative vector field. From the fundamental theorem of line integrals the inside function can therefore be written as the gradient of some scalar function

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

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

Another way to show this is by computing the closed line integral, and showing that this is always zero. From here we can use stokes theorem to show that $\nabla × (-\vec{E})=0$, and thus $\vec{E}(r)=-\nabla V$

Why is it negative? Because we define the potential function in relation to our original line integral

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It is just like a mountain slope. The higher you go the higher the potential but the force pulls you down.

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For its potential energy to increase, any particle must move to a new place in a field where a force constantly acts on it towards the direction opposite to its motion. electric field is a function of such a force on the $+1C$ charge placed at a point in the field. therefore increase in electric potential w.r.t displacement $\left( \frac{dV}{dX} \right)$ happens in a direction opposite to the electric field which is shown by the negative. just like two opposite forces have different signs.

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