Let $R$ be a right Noetherian ring which is also an algebra over $\mathbb{Q}$ ($\mathbb{Q}$ the field of rational numbers). Let $\sigma$ be an automorphism of R and $\delta$ a $\sigma$-derivation of $R$. Let further $\sigma$ be such that $a\sigma(a)\in N(R)$ implies that $a\in N(R)$ for $a\in R$, where $N(R)$ is the set of nilpotent elements of $R$. In this paper we study the associated prime ideals of Ore extension $R[x;\sigma,\delta]$ and we prove the following in this direction:
Let $R$ be a semiprime right Noetherian ring which is also an algebra over $\mathbb{Q}$. Let $\sigma$ and $\delta$ be as above. Then $P$ is an associated prime ideal of $R[x;\sigma,\delta]$ (viewed as a right module over itself) if and only if there exists ban associated prime ideal $U$ of $R$ with $\sigma(U)=U$ and
$\delta(U)\subseteq U$ and $P=U[x;\sigma,\delta]$.
We also prove that if $R$ be a right Noetherian ring which is also an algebra over $\mathbb{Q}$, $\sigma$ and $\delta$ as usual such that $\sigma(\delta(a))=\delta(\sigma(a))$ for all $a\in R$ and $\sigma(U)=U$ for all associated prime ideals $U$ of $R$ (viewed as a right module over itself), then $P$ is an associated prime
ideal of $R[x;\sigma,\delta]$ (viewed as a right module over itself) if and only if there exists an associated prime ideal $U$ of $R$ such that $(P\cap R)[x;\sigma,\delta]=P$ and $P\cap R=U$.