Electric potential Equations In our lecture, we are currently learning about electric potential and electric potential energy. 
However, I am very confused by the correct use of signs for electric potential. 
In the class, we learned that the change in potential difference is equal to $V= -Ed\cos(\theta)$. However, when solving some textbook questions, they somehow seemed to have neglected the negative sign, and just calculated the potential energy/electric field at a point to be positive. i.e. they used the equation $V=Ed\cos(\theta)$ without the negative sign. 
Anyone know why??
EDIT: Here is the question in my book I was talking about. The answer is apparently 1.5*10^6 V/m

 A: When we have two charged plates (let's assume for the sake of explanation that one is negatively charged and the other is equally positively charged) the direction electric field is from positive plate to negative plate and the picture looks something like this  
 
Now, work done on unit charge to move it from negative plate to positive plate is  what we call potential difference and therefore, we can write $$ -\int_{negative}^{positive} \mathbf E \cdot d\mathbf s = \Delta V$$ the key point here is that potential is defined as the work done against the field and negative plate is at a lower potential and positive plate is at higher potential therefore potential difference in this setup will be positive, but notice that we are moving from negative plate to positive plate therefore distance travelled will be negative 1cm (as we are moving left and in Cartesian system it is the negative direction)$$-\int_{negative}^{positive}E ~ds = 1.50 \times 10^4 V$$ $$ -E \int_{negative}^{positive}ds = 1.50 \times 10^4 V $$ $$ -E \times (-10^{-2}) = 1.50 \times 10^4 V$$ Therefore, $$ E = 1.5 \times 10^6 $$ . 
Hope this helps! If you have any doubt you're welcome to ask it in comments.
