Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems - Page 711: 21

Answer

$16.3 \mu C$

Work Step by Step

Dipole moment, $P= 1.47 D= 1.47\times3.34\times10^{-30}Cm$ Distance, $r= 52.0\times10^{-9}m$. Potential due to a dipole at a point which is on the axis of the dipole at a distance r is given by V= $\frac{1}{4\pi\epsilon_{0}}\frac{P}{r^{2}}$ $= 8.987\times10^{9}Nm^{2}/C^{2}\times\frac{1.47\times3.34\times10^{-30}Cm}{(52.0\times10^{-9}m)^{2}}$ $=1.63\times10^{-5}J/C= 16.3 \mu C$
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