Elementarily equivalence 
Suppose we have two equivalence relations $A$, $B$ s. t. $|A|=|B|$ (equipotent sets).
Is $A$ elementarily equivalent to $B$, $A \equiv B$?
My definition of elementarily equivalent is the following: $A \equiv B$ iff $Th(A)=Th(B)$.

I've tried to use the definition taking an arbitrary sentence $\varphi$ in the lexicon $\mathcal{L}$ s. t. $A \models \varphi$ or $B \models \varphi$ , also, I tried to go through equivalence classes, but I don't even know how to proceed.
I would appreciate some hint/answer, Thanks in advance!
 A: If I'm not wrong, by $|A|$ and $|B|$ you mean the cardinal of the relations themselves, not the set they're defined in, right?
(Edit: Thanks to Alex and OP comments, so it's rather cardinality of underlying set, which is the same as the cardinality of an equivalence relation on it if the set is infinite as Alex showed)
As Nagase said in comments, there's no elementary equivalence if the relations partition the set in a different (finite) numbers of classes.
For example, define on $\mathbb N$ the following equivalence relations: $A=\mathbb N\times\mathbb N$ and $B=((2\mathbb N)\times(2\mathbb N))\cup((2\mathbb N+1)\times(2\mathbb N+1))$, i.e, $xBy$ iff $x$ and $y$ have the same parity iff $x-y$ is even.
Both $A$ and $B$ are countable, but you have $A\models\forall x\forall y\, x\mathcal Ry $ (namely, there's only one equivalence class) and $B\not\models \forall x\forall y\, x\mathcal Ry$ because $(0,1)\notin B$ (in fact $B$ has two equivalence classes).
A: If you have two equivalence relations with the same finite number of classes, they'll be elementarily equivalent if and only if for each possible size of class (an element of $\mathbb{N} \cup \{\infty\}$, we don't distinguish infinite cardinals) they have the same number of classes with that size.
e.g say $A$ has two classes, one with 2 elements, the other with infinitely many elements. Say $B$ has two classes, one with 3 elements, the other with infinitely many elements. Then $A$ and $B$ are not elementary equivalent since the formula $\varphi := \exists x \exists y \ \Big(x \neq y \wedge \forall z\ \ (xEz \to (z = y \vee z = x)\big)\Big)$ holds in $A$, but not in $B$.
