University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 24 - Capacitance and Dielectrics - Problems - Exercises - Page 810: 24.18

Answer

(a) $Q = 115.2 \mathrm{~\mu C}$ (b) $V_1 = 38.4 \,\text{V}$ and $V_2 = 23.04\,\text{V}$

Work Step by Step

(a) $C_1$ and $C_2$ are in series, which means they have the same charge and their equivalent capacitance is calculated by $$\dfrac{1}{C_{12}} = \dfrac{1}{C_{1} } + \dfrac{1}{C_{2} } = \dfrac{1}{3\mathrm{~\mu F} } + \dfrac{1}{5 \mathrm{~\mu F} } = 0.533 $$ Take the reciprocal $$C_{12} = 1.8 \mathrm{~\mu F} $$ Therefore, the charge on each plate is given by $$Q = C_{12} V = (1.8 \mathrm{~\mu F} ) (64 \,\text{V}) = \boxed{115.2 \mathrm{~\mu C}}$$ (b) The potential in the series connection is different for each capacitor. For $C_1$ the potential is given by $$V_1 = \dfrac{Q}{C_1} = \dfrac{115.2 \mathrm{~\mu C}}{3\mathrm{~\mu F}} = 38.4 \,\text{V}$$ For $C_2$ the potential is given by $$V_2 = \dfrac{Q}{C_2} = \dfrac{115.2 \mathrm{~\mu C}}{5\mathrm{~\mu F}} = 23.04 \,\text{V}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.