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: 24.44


(a) $Q = 5 \times 10^{-9} \mathrm{~ C}$ (b) $E = 1.2 \times 10^{4} \,\text{N/C}$ (c) $E_i = 2.5 \times 10^{4} \,\text{N/C}$

Work Step by Step

(a) First, let us find the capacitance by $$C = \dfrac{K \epsilon_o A}{d} = \dfrac{(2.1) (8.85 \times 10^{-12} \,\text{F/m}) (0.0225 \mathrm{~m^2})}{ 1 \times 10^{-3} \,\text{m}} = 418 \times 10^{-12} \,\text{F}$$ After calculating the capacitance we could determine the charges where the capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential of the battery and it is given by $$Q = CV_{ab}$$ Where $V_{ab} $ is the potential between the two plates. Let us substitute the values $C$ and $V_{ab}$ to get the charge on each plate $$Q = CV_{ab} = (418\times 10^{-12} \,\text{F})(12 \,\text{V}) = \boxed{5 \times 10^{-9} \mathrm{~ C}}$$ (b) We integrate equation 24.23 for $A$ to determine the electric field by \begin{gather*} \oint K \vec{E} \cdot d\vec{A} = \dfrac{Q}{\epsilon_o} \\ K E A = \dfrac{Q}{\epsilon_o} \\ E = \dfrac{Q}{\epsilon_o K A}\\ E = \dfrac{5 \times 10^{-9} \mathrm{~ C}}{(8.85 \times 10^{-12} \,\text{F/m}) (2.1) (0.0225 \mathrm{~m^2})} \\ \boxed{E = 1.2 \times 10^{4} \,\text{N/C}} \end{gather*} (c) The electric field will be calculated with the same steps but without $K$ and could be calculated by \begin{gather*} E_i = \dfrac{Q}{\epsilon_o A}\\ E_i = \dfrac{5 \times 10^{-9} \mathrm{~ C}}{(8.85 \times 10^{-12} \,\text{F/m}) (0.0225 \mathrm{~m^2})} \\ \boxed{E_i = 2.5 \times 10^{4} \,\text{N/C}} \end{gather*}
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.