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: 24.42


(a) $K = 1.8$ (b) $V = 2 \,\text{V}$ (c) $E = 1000 \,\text{V/m}$

Work Step by Step

(a) The charges on the plates are directly proportional to the dielectric constant $K$ therefore, we could get $K$ by $$K = \dfrac{Q_K}{Q} = \dfrac{ 45 \,\text{pC} }{ 25 \,\text{pC} } =\boxed{ 1.8 } $$ (b) The battery is kept connected therefore, the potential difference will be constant and we could calculate it before inserting the dielectric material $$V = \dfrac{Q}{C} = \dfrac{25 \,\text{pC}}{12.5 \,\text{pF}} = \boxed{2 \,\text{V}}$$ (c) To find the electric field we should find the separated distance $d$ by \begin{align} d = \dfrac{\epsilon_o A }{C} = \dfrac{(8.85 \times 10^{-12} \,\text{F/m}) \pi( 0.03 \,\text{m})^2 }{12.5 \times 10^{-12} \,\text{F}} = 0.002 \,\text{m} \end{align} The electric field is calculated by $$E = \dfrac{V}{d} = \dfrac{2 \,\text{V}}{0.002 \,\text{m}} = \boxed{1000 \,\text{V/m}}$$
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