For capacitors in series, the equivalent capacitance is for two capacitors:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}\frac{{\mathbf{C}}_{\mathbf{1}}{\mathbf{C}}_{\mathbf{2}}}{{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}}}$

For capacitors in parallel, the equivalent capacitance is:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{n}}}}$

Charge:

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

C_{2} and C_{3} are in parallel.

C_{23} = C_{2} + kC_{3}

C_{1} and C_{23} are in series.

C_{123} = C_{1}(C_{2} + kC_{3})/(C_{1} + C_{2} + kC_{3})

The switch S is closed for a long time and charges the capacitors. There is a dielectric k in C3. Find the charge and potential on each capacitor.

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