Paradox for solving E in terms of potentials $$ \int E\cdot dl  =  - \int \frac{\partial B}{\partial t} \cdot dA  $$
If $$\nabla ×A=B$$
$$ \int E\cdot dl  =-  \int \frac{\partial (\nabla ×A)}{\partial t} \cdot dA  $$
$$ \int E\cdot dl  =- \int \nabla\times\frac{\partial A}{\partial t} \cdot dA  $$
Invoking Stokes theorem gives:
$$ \int E\cdot dl  =\int \frac{-\partial A}{\partial t} \cdot dl$$
$$E=-\frac{\partial A}{\partial t}$$
which differs by
$$-\nabla V$$
from the standard result you would obtain by evoking stokes theorem on the $E$ line integral, changing to differential form and then substituting the magnetic vector potential and then solving for $E$
Where have I gone wrong? as the derivative of the curl of $A$ is the same as the curl of the derivative of $A.$ At least I think, and this is also taken to be true in the actual derivation.
(Edit: I have solved the paradox I believe in the comment below)
for a closed loop
$$\int \nabla V \cdot dl = 0$$
$$ \int E.dl  =\int \frac{-\partial A}{\partial t} \cdot dl - \int \nabla V\cdot dl$$
$$ \int E\cdot dl  =\int \left( \frac{-\partial A}{\partial t}-\nabla V \right) \cdot dl$$
$$E = -\nabla V - \frac{\partial A}{\partial t} $$
So I can simply not cross of the ${}\cdot dl$ and set the integrand equal, for a line integral I need to add the gradient of a scalar function.
 A: Upon further inspection I think I have solved the problem. This is a closed line integral, so the integral can have a gradient of a scalar function added on without changing the result via the fundamental theorem of gradients so I can simply not cross of the line integral, I have to add the gradient of a scalar function when I solve for the fields. I guess I have learned i need to be more carefull about simply crossing off the .dl and setting the integrand to be equal when in reality this is a special case where grad v is zero
for a closed loop
$$\int \nabla V .dl = 0$$
$$ \int E.dl  =\int-\partial A / \partial t.dl - \int \nabla V .dl$$
$$ \int E.dl  =\int(-\partial A / \partial t-\nabla V ) .dl$$
$$E = -\nabla V - \partial A / \partial t $$
So I can simply not cross of the .dl and set the integrand equal, for a line integral I need to add the gradient of a scalar function.
when ever using stokes theorem to change from a surface integral to a line integral, you must add the gradient of a scalar function
