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 $\nabla × F = 0$

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)

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)