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 the figure. Show that the electron will fail to strike the upper plates if

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Solution

Given:
Potential difference across the plates of the capacitor = V
Separation between the plates = d
Magnetic field intensity = B
The electric field set up between the plates of a capacitor, $E = \frac{V}{d}$
The force experienced by the electron due to this electric field, $F = \frac{e V}{d}$
$\Rightarrow a = \frac{F}{m} = \frac{e V}{m_{e} d}$ ,
where e = charge of the electron and m_e = mass of the electron
Using v² = u²+ 2as and substituting the value of a, we get:
$v^{2} = 2 \times (\frac{e V}{m_{e} d}) \times d \Rightarrow v = \sqrt{\frac{2 e V}{m_{e}}}$
The electron will move in a circular path due to the given magnetic field. Radius of the circular path,
$r = \frac{m_{e} v}{e B}$
And the electron will fail to strike the upper plate only when the radius of the circular path will be less than d,
$i . e . d > r \Rightarrow d > \frac{m_{e}}{e B} \times \sqrt{\frac{2 e V}{m_{e}}} \Rightarrow d > \sqrt{\frac{2 m_{e} V}{e B^{2}}}$
Thus, $d > {(\frac{2 m_{e} V}{e {B_{0}}^{2}})}^{\frac{1}{2}}$

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