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: 24.8


(a) $d$ = 10 mm and $r$ = 4.24 cm (b) $Q = 5 \times 10^{-10} \mathrm{~ C}$

Work Step by Step

(a) We want to calculate the dimensions and the separation distance $d$. To determine the dimensions we must find the radius from the area. First, let us find the separated distance where the electric field between the two plates depends on the distance by $$d = \dfrac{V}{E}$$ Where $V$ is the potential. Substitute the values of $V$ and $E$ to find $d$ $$d = \dfrac{1.00 \times 10^{2} \,\text{V} }{1.00 \times 10^{4} \,\text{N/C}} = 10 \,\text{mm}$$ We could use equation 24.2 in the textbook to determine the area $A$ in the form \begin{equation} A = \pi r^2= \dfrac{Cd}{\epsilon_o } \tag{1} \end{equation} Let us substitute the values of $\epsilon_o, C$ and $d$ into equation (1) to get the value of $r$ \begin{align} r &= \sqrt{\dfrac{Cd}{\pi \epsilon_o }}\\ & = \sqrt{\dfrac{(5\times 10^{-12} \,\text{F})(0.01 \,\text{m})}{\pi (8.85 \times 10^{-12} \,\text{F/m})}}\\ &= 4.24 \,\text{cm} \end{align} (b) The capacitance between the two plates occur due to the accumulation of the charges on the plates and it is given by $$Q = CV_{ab}$$ Substitute the values $C$ and $V_{ab}$ to get the charge on the plate $$Q = CV_{ab} = (5 \times 10^{-12} \,\text{F})(1.00 \times 10^{2}\,\text{V}) = 5 \times 10^{-10} \mathrm{~ C}$$
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.