Question

An electron is emitted with negligible speed from the negative plate of a parallel plate capacitor charged to a potential difference $V$. The separation between the plates is $d$ and a magnetic field $B$ exists in the space as shown in figure. Show that the electron will fail to strike the upper plate if
$d>{\left(\frac{2m_{e}V}{e{B}_0^2}\right)}^{\frac{1}{2}}$.

Answer 1

The force experienced first is due to the electric field of the capacitor

$E=\frac{V}{d}$ and $F=eE$

$\therefore$ acceleration

$a=\frac{eE}{m_e}$

where $m_e$ = mass of electron

using

${\vartheta }^2=u^2+2as$

${\vartheta }^2=0+2\left(\frac{eE}{m_e}\right)\ast d$

${\vartheta }^2=\frac{2eVd}{d{m}_e}$

$\vartheta =\sqrt{\frac{2eV}{m_e}}$

now, the electron will fail to strike the upper plate only when d is greater than radius of the arc thus formed.

$d>r$

where

$r=\frac{m_e\ast \vartheta }{qB}=\frac{m_e\sqrt{\frac{2eV}{m_e}}}{eB}$

$d>\frac{\sqrt{2e{m}_{e}V}}{eB}$

$d>{\left(\frac{2m_{e}V}{e{B}^2}\right)}^{\frac{1}{2}}$