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?
Electrostatics – Why Is the Electric Field the Negative Gradient of Potential Energy?
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One thing necessary about Potential. It's simply the work done in moving an electric charge against the electrical force. Thus, The negative sign actually says that the work done is against the electric force (either attraction or repulsion). Or in other words, electric potential decreases in the direction of electric field. Hence, you're quite right about the concept.
This link is explains in an imagery way.
I believe you are confusing the work done by the electric field with the work done on the particle.
By the electric field: The radial force of the electric field is always pointing outwards, and the displacement of the charge in this case is going inward. Thus, the integral you've specified will be negative. That is, the work done by the electric field is negative.
On the Particle: The force that must be applied to the charge (ignoring gravity, relativity, induced vector-potential, etc.) to get it from infinity to the radial distance $r$ is, at the very least, equal and opposite to the force of the electric field acting on the charge. Thus, it is always pointing inwards. Given that the charge is going inwards, the work integral will come out positive.
Best Answer
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$$d \phi=\left(\vec {i}\frac{\partial \phi}{\partial x}+\vec {j}\frac{\partial \phi}{\partial y}+\vec {k}\frac{\partial \phi}{\partial z}\right).\left(dx\ \vec{i}+dy\ \vec{j}+dz\ \vec{y}\right)=\vec{\text{grad}}\ \phi.d\vec{r}$$