Voltage divider rule:

$\overline{){{\mathbf{V}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{\epsilon}}\frac{{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{R}}_{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{\epsilon}}\frac{{\mathbf{R}}_{\mathbf{1}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{R}}_{\mathbf{2}}}}$

The charge stored on a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

When the capacitor is fully charged, there is no current through R_{3}. The voltage across the capacitor is equal to the voltage across R_{2}.

V_{C} = V_{2} = ε[R_{2}/(R_{1} + R_{2})] = 42.0[6.00/(6.00 + 8.00)] = 18 V

The capacitor in the figure shown is initially uncharged. The switch is closed at t = 0.

What is the final charge on the capacitor?

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