Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 17 - Electric Potential - Problems - Page 498: 66

Answer

$E=3.21\times10^5V/m$

Work Step by Step

For this problem, we will assume no vertical acceleration when the electron is accelerated horizontally and no horizontal acceleration when in the uniform electric field. Therefore, horizontal component of velocity, $v_x$ is entirely dependent on the voltage of $9kV$ When in the electric field, it gives the electron a vertical impulse of $Ft=eEt=mv_y$ $y=11cm$, $x=22cm$, $t_y=t_x$, so $v_y=\frac{v_x}{2}$ $\frac{mv_x^2}{2}=eV$ because electric energy can be converted to kinetic energy $E=\frac{mv_y}{et}=\frac{mv_xv_y}{ed}=\frac{mv_x^2}{2ed}=\frac{V}{d}=\frac{9\times10^3V}{2.8\times10^{-2}m}$$=3.21\times10^5V/m$
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