Why do electric field lines point in the direction of decreasing electric potential? Why do electric field lines point in the direction of decreasing electric potential? 
I came across this sentence in my school book but am trying to understand this ever since. 
I know that dV=-E.dr 
But how do I get to the above conclusion from this? 
 A: It is a convention that field lines point from positive to negative. 
Now because they point from positive to negative we get 
$$E = -{dV\over dr}$$
(in spherical symmetry).
If you have a positive charge at the origin then the potential at the origin is positive and it decreases as you move away from the origin - thus $dV \over dr$ is negative because $dV \over dr$ is the rate of change of potential with $r$ and as $r$ increases $V$ decreases... and so from the equation above $E$ is positive, which means it points in the positive $r$ direction away from the positive charge at the origin.
If you have a negative charge at the origin then the potential at the origin is negative and it increases (approaches zero) as you move away from the origin - thus $dV \over dr$ is positive and $E$ is negative, which means it points in the negative $r$ direction towards the origin where the negative charge is.
A: The potential, $\phi$, is defined as the the potential energy, $U$, of a system of charges, $Q$ and a unit charge. The potential energy of a system of charges is defined to be the negative of the work done by the electric field of one charge, $Q$, on another charge $q$, as $q$ is moved from infinitely far away to some distance $r$ from $Q$.  If $q$ is a unit charge, the negative of the work done will be the potential due to the field of $Q$ at point $r$. Now, let's do the math:
$$ W_{Qq} = \int \vec{F}_{Qq}\cdot d\vec{r} = \int q\vec{E}_{Q}\cdot d\vec{r}$$
For $q=1$ we can write
$$ \phi_Q=U_{Q1}=-W_{Q1}(\infty\rightarrow r)$$
$$ \phi_Q(r)= - \int_{infty}^r \frac{kQ}{r^2}\hat{r}\cdot dr\hat{r} $$
$$ \phi_Q(r) = \frac{kQ}{r} $$
If $Q$ is positive, the electric field is defined to point away from the charge by $\vec{E}=\vec{F}/q$. The result which we obtained here for the potential shows that the potential is positive and decreases with increasing $r$.
If $Q$ is negative, the electric field is defined to point toward the charge. The potential is negative and becomes less negative (increases) with increasing $r$.
