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:

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

A 0.50-μF and 1.4-μF capacitor (C_{1} and C_{2}, 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.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solving Capacitor Circuits concept. You can view video lessons to learn Solving Capacitor Circuits. Or if you need more Solving Capacitor Circuits practice, you can also practice Solving Capacitor Circuits practice problems.