## Fundamentals of Physics Extended (10th Edition)

Published by Wiley

# Chapter 25 - Capacitance - Problems - Page 739: 3a

#### Answer

$144\ pF$

#### Work Step by Step

It is given that: Radius of each plate$=8.20\ cm = 8.20 \times 10^{-2} m$ Separation between the plates $d = 1.30 mm = 1.30\times 10^{-3}\ mm$ Area of the circle $A = \pi r^2 = \pi (8.20 \times 10^{-2}\ m )^2 = 2.11\times 10^{-2}\ m^2$ The formula for the capacitance having area $A$ and distance separation $d$ is; $C =\frac {\epsilon_o A}{d}$ $C =\frac {(8.85\times 10^{-12}\ c^2/N.m^2)(2.11\times 10^{-2}\ m^2)}{1.30\times 10^{-3}\ m}$ $C = 144\times 10^{-12} F$ $C = 144\ pF$

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