Q.
An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma$. Electric intensity in the space between the plates is $E$. A uniform magnetic field $B$ also exists in that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$ without any change in direction. The time taken by the electron to travel a distance $l$ in that space is
A) $\frac{\sigma l}{\varepsilon_{0}B}$
B) $\frac{\sigma B}{\varepsilon_{0}l}$
C) $\frac{\varepsilon_{0}lB}{\sigma}$
D) $\frac{\varepsilon_{0}l}{\sigma B}$
Explanation
(c) Force on the electron due to the electric field $E$ is $F_{E} = ( - e)E$
Force on the electron due to the magnetic field $B$ is $F_{B} = ( - e)vB$
The electron will move in the fields undeflected, if these two forces are equal and opposite
$eE = evB$ or $v = \frac{E}{B}$
Electric field between the plates is $E = \frac{\sigma}{\varepsilon_{0}}$
$\therefore v = \frac{\sigma}{\varepsilon_{0}B}$
The time take by the electron to travel a distance $l$ in the space is $t = \frac{l}{v} = \frac{l}{\frac{\sigma}{\varepsilon_{0}B}} = \frac{l\varepsilon_{0}B}{\sigma}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.