Explanation

(a) We know that in steady state the capacitor behaves like as an open circuit $ie$, capacitor will not pass the current.
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So, the potential difference across the capacitor = 45 V
Hence , the final charge on the capacitor is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = CV$
Here , $C = 20\mu F,\ \ \ \ \ \ \ \ V = 45\ V$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = 20 \times 10^{- 6} \times 45$
or $\ q = 900 \times 10^{- 6}\ $
or $\ q = 9 \times 10^{- 4}C\ $