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{$\vec E$ - Electric Field Vector ; $x^2$ is $x$ raised to the power $2$}

When finding the Potential Difference between two points in a non uniform electric field, the equation of $\vec E $ given in the question is something like this

$$\vec E = 2xydx + (x^2)dy$$

Upon integrating (to find the potential difference) the $x$ and $y$ components separately, we get $(x^2)y$ as the result in both the cases and thus the potential is $(x^2)y$ and the function in $\vec E$ is an exactly differential.

The book I am referring says that the any function in $\vec E$ will be exactly differentiable as $\vec E$ is an conservative field and exact differentiability is a property of all conservative fields.

My question is : What if the question is a function in $\vec E$ which is not an exactly differential? If it is, then does it mean that such an $\vec E$ cannot exist? and thus is the potential difference also zero?

Also, Why are electric field functions an exact differential? I have some faint understanding relating to the fact that electrostatic forces are conservative forces and thus the work done by them does not depend on the path followed. I am not really satisfied with this explanation of mine though.

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In electrostatics, the electric field is the gradient of some potential $V(\vec{r})$. It's fairly straightforward to prove that if a continuously-differentiable field is the gradient of a scalar field, then that vector field is an exact differential.

Consider the two-dimensional case for clarity of notation. We have an electric field $\vec{E}=M\hat{x}+N\hat{y}$, and we also know that $\vec{E}=-\nabla V$, so $M=-\frac{\partial V}{\partial x}$ and $N=-\frac{\partial V}{\partial y}$. Therefore:

$$\frac{\partial M}{\partial y}=-\frac{\partial^2 V}{\partial y\partial x}$$

$$\frac{\partial N}{\partial x}=-\frac{\partial^2 V}{\partial x\partial y}$$

We have assumed that $\vec{E}$ is continuously differentiable, so the partial derivatives are continuous, meaning that by Clairaut's theorem, the two right-hand sides are equal. Therefore, $\vec{E}$ is an exact differential of $V$.

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