Why does error propagation in CBC mode encryption affect two blocks?
Encryption process in CBC mode is performed as
\begin{align}
C_1 &amp;= Enc_k(P_1 \oplus IV)\\
C_i &amp;= Enc_k(P_i \oplus C_{i-1}),\;\; 1 &lt; i \leq nb,
\end{align}
where $nb$ is the number of blocks.

CBC is designed for chaining, therefore while encryption, if there is one block error at position $i$, it will affect the rest of ciphertext. This also can be seen from below if write the equation of encryption of $j$-th block.

$$C_j = Enc_k(P_j \oplus Enc_k(P_{j-1} \oplus \cdots Enc_k(P_1 \oplus IV)\cdots)).$$ If $i &lt; j$  then the $j$-th plaintext block will be affected.

However, while decryption of the ciphertext, that is encryption of the plaintext that contains the errored block, you will get only one block with error, the errored plaintext block. Or simply, you will get what you encrypted if there is no error/corruption on the ciphertext blocks.

Note: if there is an error on one of the ciphertext blocks, it can affect one or two plaintext blocks depending on the index of the ciphertext block. One block if the last ciphertext block is corrupted, two blocks otherwise, see bit flipping attack to visiualize. The question is on the error on the plaintext.