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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?

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Consider the following:
The point $P$ in an electric field has an electric potential of $12\ V$. Where this point is in the field depends on the field itself. The two fields shown below have different maxima so point P is at a different point in each, but the electric potential at $P$ is still $12\ V$ in each field.

Point P in an electric field

The charge $q$ at point $P$ has a potential energy of $E=qV$ in each field. Because the electric potential at $P$ is the same in both fields, the potential energy $E$ for the charge $q$ is the same in both fields. At this point we don't need to consider whether work was done on $q$ or by $q$, we can just take this as a starting point and say that the charge $q$ is at point $P$, and if $q=-2\ C$ then $E=qV=(-2)(12)=-24\ J$.

Charge q at point P in an electric field

Whether we want to then consider how $q$ got to $P$, or what happens next with $q$ moving from $P$, we need to understand whether work is being done on $q$ or by $q$.

For a positive charge $+q$, to move from a lower electric potential to a higher electric potential, work needs to be done on $q$ because the potential energy is increasing as the electric potential increases, so we need to add energy to $q$. Conversely, to move from a higher electric potential to a lower electric potential, work is done by $q$ because the potential energy of $q$ decreases as it moves to a lower electric potential.

On the other hand, for a negative charge $-q$, to move from a lower electric potential to a higher electric potential, work is done by $q$ because the potential energy is decreasing (becoming more negative) as the electric potential increases, e.g. from $E=qV=(-2)(8)=-16\ J$ to $E=qV=(-2)(12)=-24\ J$. That is, $q$ is losing energy. And conversely, to move from a higher electric potential to a lower electric potential, work must be done on $q$ because the potential energy of $q$ increases (becoming more positive) as it moves to a lower electric potential, e.g. from $E=qV=(-2)(12)=-24\ J$ to $E=qV=(-2)(8)=-16\ J$.

And if $q$ moves, or is moved, sideways in the field then no work is done because the electric potential does not change and so the potential energy of $q$ does not change.

Charge q 'has work done' or 'does work' to arrive at point P in an electric field

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  • $\begingroup$ Excellent answer Mick - can I ask, if the charge is moving due to a force acting on it, using the above information, can one calculate the electric flux (assuming this were a $3$-d model imposed on a $2$-d page!). $\endgroup$ – Kevin Aug 31 '18 at 14:07
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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.

For a negative charge, the potential at infinity will actually be higher than at point P.

This should make sense, considering that a negative charge would be attracted to P and, therefore, the work, "required" to move it from infinity to P, would have to be negative.

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