I really confused from the relation of potential difference and electric field.

as you know the relation is :
$$V{ab}=-\int_{b}^a{\vec{E}\cdot\vec{dr}}$$ 

now let's look on a solid sphere uniformly charged $q$ and let's find the potential difference $V{ab}$ while $b\to\infty$

the electric field of solid sphere is in the $\vec{r}$ direction hence if we are going from $b$ to $a$ we are going **against** the electric field (because $b>a$).

and we getting:
$$V{ab}=-\int_{b}^a{\vec{E}\cdot\vec{dr}} =-\int_{b}^a E*dr*cos(\pi) = 
\int_{b\to\infty}^a \frac{kq}{r^2}*dr = -\frac{kq}{a}<0 $$ 

now the results make no sense because it means that $V{a}<V{b}$ which means that the electric field of the solid sphere is in the $-\vec{r}$ direction and this is wrong.

what I'm doing wrong? i saw many solutions of exercises wich using this relation and it seems that each one of them just solving the integral without considering the dot product.

I hope it was clear, thanks a lot!