For a conservative electric field, we can always say that $$\oint \vec E\cdot\vec {dl} = 0$$
Take this scenario for example
Here $E_1$ and $E_2$ are two different electric fields. If I find
$\oint \vec E\cdot\vec {dl}$ from $A$ to $A$(loop), then I know it is $0$, because the potential difference between $A$ and $A$ will always be zero (since they are the same point).
Now, here only vertical lines (PQ and RS) would be nonzero.
So,
$$\oint \vec E\cdot\vec {dl} = 0$$
$$\int_P^Q \vec E_1\cdot\vec {dl} + \int_S^R \vec E_2\cdot\vec {dl} = 0$$
$$E_1\cdot l – E_2\cdot l = 0$$
$$E_1 = E_2$$
But then, it contradicts my assumption of $E_1$, $E_2$ being different.
It gives rise to two possibilities:
- The potential difference between $A$ and $A$ is not zero.
- You can never create two different electric fields, because if you do, I could always draw a closed loop encapsulating both the fields and prove that the fields are same.
So what is going wrong here?
Best Answer
Without a magnetic field existing, $\oint{\vec{E}\cdot d\ell=0}$ is always true.
Your mistake is that you don't consider the edge effect of a parallel plate capacitor. As the picture you can see,
when you approach the edge of the parallel plates, the horizontal component of the electric field cannot be neglected. Counting this effect in, you can have two different electric fields, and the law of a conservative electric field is still true.