## Physics (10th Edition)

Published by Wiley

# Chapter 19 - Electric Potential Energy and the Electric Potential - Problems - Page 539: 47

5.3

#### Work Step by Step

Capacitance C= 7.0$\mu F$= $7.0\times10^{-6}F$ Area A= 1.5 $m^{2}$ Plate separation d= $1.0\times10^{-5}m$ $\epsilon_{0}= 8.854\times10^{-12} F/m$ Recall that C= $\frac{K\epsilon_{0}A}{d}$. Thus we get that the dielectric constant K= $\frac{Cd}{\epsilon_{0}A}$ $= \frac{(7.0\times10^{-6}F)(1.0\times10^{-5}m)}{(8.854\times10^{-12}F/m)(1.5m^{2})}\approx5.3$

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