College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 19 - Problems - Page 755: 45


$\frac{q}{m} = \frac{E^2}{2~B^2~\Delta V}$

Work Step by Step

The particle has a kinetic energy of $q~\Delta V$. We can find an expression for the velocity $v$: $K = q~\Delta V$ $\frac{1}{2}mv^2 = q~\Delta V$ $v^2 = \frac{2~q~\Delta V}{m}$ $v = \sqrt{\frac{2~q~\Delta V}{m}}$ If the particle passes straight through, then the net force on the particle is zero. If the net force is zero, the magnitude of the force from the magnetic field must be equal in magnitude to the force from the electric field. We can find an expression for $\frac{q}{m}$: $q~v~B = E~q$ $\sqrt{\frac{2~q~\Delta V}{m}} = \frac{E}{B}$ $\frac{2~q~\Delta V}{m} = \frac{E^2}{B^2}$ $\frac{q}{m} = \frac{E^2}{2~B^2~\Delta V}$
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