For capacitors in series, the equivalent capacitance is:

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

Charge:

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

**(a)**

Equivalent capacitor:

1/C_{eq} = 1/4.0 + 1/2.0 + 1/1.0 = 7/4

**(a)** What is the equivalent capacitance for the circuit of the figure?**(b)** What is the charge of the capacitor 4.0 *μ*F?**(c)** What is the charge of the capacitor 2.0 *μ*F?**(d)** What is the charge of the capacitor 1.0 *μ*F?

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