Q.Find the Q point for the circuit of figure 2.1.11 considering transistor to be of silicon.

The Q point of a voltage divider bias circuit is found from the divider voltage $V_2$, then $I_{CQ} = (V_2 - V_{BE})/R_E$ and $V_{CEQ} = V_{CC} - I_{CQ}(R_C + R_E)$, using $V_{BE} = 0.7\ \text{V}$ for silicon. …

[!TLDR]

The operating point is $Q(10.8\ \text{V},\ 0.39\ \text{mA})$.

The Q point (quiescent/operating point) is the zero-signal point on the load line, with coordinates $Q(V_{CEQ}, I_{CQ})$. For voltage divider bias it is computed from the potential-divider output voltage across $R_2$.

Divider voltage:

Quiescent collector current (with $I_E \approx I_C$ and silicon $V_{BE} = 0.7\ \text{V}$):

Quiescent collector-to-emitter voltage:

$V_{CEQ} = V_{CC} - I_{CQ}(R_C + R_E) = 12 - (0.39\times10^{-3})(2\times10^{3} + 1\times10^{3}) = 12 - 1.17 = 10.8\ \text{V}$ …