Equivalent capacitance for two capacitors connected in series is:

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

The potential difference across a capacitor is given as:

$\overline{){\mathbf{\u2206}}{\mathbf{V}}{\mathbf{=}}\frac{\mathbf{Q}}{\mathbf{C}}}$

**A)**

The capacitors are connected in series.

Therefore, the total charge Q on each capacitor is the same.

A 0.50-μF and a 1.4-μF capacitor (C1 and C2, respectively) are connected in series to a 7.0-V battery.

A) Calculate the potential difference across each capacitor

B) Calculate the charge on each capacitor

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