A positive charge $q$ is placed somewhere. So electric lines of forces will be radially outward from this charge. While defining potential at a distance $r$ from $q$, we say it is the work done to bring a unit positive charge from infinity to that distance. I have a confusion here. Since point of infinity can be anywhere outside the electric field,there are lots of points of infinity and by bringing a unit positive charge we can bring that charge in any way we want,we can take a zigzag path or an absurd trajectory,or we go some distance forward and then go more distance backward i mean the path can be however we want. So the way we will bring the charge is not defined properly. How do we then deduce the formula for potential if we are bringing the charge in a curved path?

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    $\begingroup$ Concluding that “the way we will bring the charge is not defined properly” from “the path can be however we want” isn’t reasonable. The property which makes the path irrelevant is $\nabla\times\vec E=0$. $\endgroup$
    – Newbie
    Jan 12, 2022 at 7:35

1 Answer 1


As others have stated. The reason that the path taken is IRRELEVANT in electrostatics, is that the electric field can be written as $-\nabla V$. When a function can be written as the gradient of a scalar function, the line integral is path independant. This is an example of a Conservative field, where $\nabla × \vec{E}= 0$

For a formal proof, see my answer here, Proving if a force is conservative and non-conservative

Solving the equation

$\vec{E} = -\nabla V$

For V, yields the standard potential function.( can also be done using maxwells equations themself)


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