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 811: 24.34


(a) $Q_o = 0.36 \mathrm{~n C}$ (b) $Q_K = 0.97 \mathrm{~n C}$

Work Step by Step

(a) the capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential and for the air, it is given by $$Q_o = C_oV$$ Where $V_{ab} $ is the potential between the two plates and is calculated by $$V = Ed = ( 3 \times 10^{4} \,text{V/m})(0.0015 \,\text{m}) = 45 \,\text{V}$$ Let us substitute the values $C$ and $V$ to get the charge on each plate $$Q_o = C_oV = (8\times 10^{-12} \,\text{F})(45 \,\text{V}) = \boxed{0.36 \mathrm{~n C}}$$ (b) After the dielectric material with a dielectric constant is added $K = 2.70 $ the charges will be given by $$Q_K = K C_oV = 2.70 (8\times 10^{-12} \,\text{F})(45 \,\text{V}) = \boxed{0.97 \mathrm{~n C}}$$
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