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 809: 24.11


(a) $C$ = 15 pF (b) $r_a$ = 3 cm (c) $E = 3.3 \times 10^{4} \,\text{N/C}$

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 difference and it is given by $$C = \dfrac{ Q}{V_{ab}}$$ Where $V_{ab} $ is the potential between the two plates. Let us substitute the values $Q$ and $V_{ab}$ to get the capacitance $$C = \dfrac{ Q}{V_{ab}} = \dfrac{3.30 \times 10^{-9} \,\text{C}}{220 \,\text{V}} = 15 \,\text{pF}$$ (b) The capacitance for a spherical capacitor is calculated by \begin{gather*} C = 4 \pi \epsilon_o \dfrac{r_a r_b}{r_b - r_a}\\ C = 4 \pi \epsilon_o \dfrac{1 }{\dfrac{1}{r_a} - \dfrac{1}{r_b}}\\ \dfrac{1}{r_a} - \dfrac{1}{r_b} = \dfrac{4 \pi \epsilon_o }{C} \\ \dfrac{1}{r_a} = \dfrac{4 \pi \epsilon_o }{C} + \dfrac{1}{r_b} \end{gather*} Substitute for $C, \epsilon_o$ and $r_b$ \begin{align*} \dfrac{1}{r_a} &= \dfrac{4 \pi \epsilon_o }{C} + \dfrac{1}{r_b} \\ &= \dfrac{4 \pi (8.85 \times 10^{-12} \,\text{F/m}) }{15 \times 10^{-12} \,\text{F}} + \dfrac{1}{0.04 \,\text{m}}\\ &= 32.4 \mathrm{~m^{-1}} \end{align*} Now take the reciprocal of the answer $$r_a = \dfrac{1}{32.4 \mathrm{~m^{-1}}} = 3 \mathrm{~cm}$$ (c) The electric field of the sphere is given by Gauss's law in the form \begin{align*} E = \dfrac{1}{4\pi \epsilon_o} \dfrac{Q}{r_a^2} \end{align*} Substitute to get $E$ \begin{align*} E = \dfrac{1}{4\pi \epsilon_o} \dfrac{3.30 \times 10^{-9} \,\text{C}}{( 0.03 \mathrm{~m})^2}= 3.3 \times 10^{4} \,\text{N/C} \end{align*}
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