{$\vec E$ - 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.


1 Answer 1


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 $V$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.