From what I understand the electric potential at a point $P$ is the amount of work done per unit charge $q$ in bringing that charge $q$ to the point $P$. Now consider the following question: The potential at the point $P$ is $12$ V and a charge of $3C$ is placed there, then the potential energy is simply $12\times 3 = 36$J.
Here is my confusion, for the point $P$ to have a potential of $12$V, it implies that the charge being moved which is $3C$ is moving from a lower to a higher potential before it got to the point $P.$ So perhaps near the point $P$ there is a higher charge involved say $10C.$ But if $P$ were really near a high positive charge $10C$, then to bring a $3C$ charge from very far away to the point $P$, then since the electric field lines from $10C$ is greater than $3C$, ultimately work has to be done in overcoming that field and hence is the reason why the work done is positive? Because we need to do work?
Also consider the same situation as above but now we are moving a charge of $-2C$ from far away to the point $P$. Again near point $P$ or at point $P$ it is of positive potential of $12V$ which implies to move $-2C$ to that point, it must be moving from a lower to a high potential. But if it really is at a higher potential near $P$ say of $1C$, then the field lines are from $1C$ towards the $-2C$ which implies work has to be done to move the $-2C$ from far away to the point $P$, but then the energy required is $-2(12) = - 24$J which implies energy was lost during moving $-2C$ from far to the point $P.$
Where in my understanding am I going wrong?