# Problem: A 0.50-μF and 1.4-μF capacitor (C1 and C2, respectively) are connected in series to a 9.0-V battery. Part A. Calculate the potential difference across each capacitor. Express your answers using two significant figures separated by a comma.Part B. Calculate the charge on each capacitor assuming the two capacitors are in parallel.

###### FREE Expert Solution

For two capacitors in series, the equivalent capacitance is given by:

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

Potential difference:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{Q}}{\mathbf{C}}}$, where Q is charge and C is capacitance.

Charge, Q:

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

100% (325 ratings) ###### Problem Details

A 0.50-μF and 1.4-μF capacitor (C1 and C2, respectively) are connected in series to a 9.0-V battery.

Part A. Calculate the potential difference across each capacitor. Express your answers using two significant figures separated by a comma.

Part B. Calculate the charge on each capacitor assuming the two capacitors are in parallel.

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