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 dand 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 `d > ((2m_eV)/(eB_0^2))^(1/2)`

Answer 1

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 = V/d`

`⇒ a = F/m = (eV)/(M_ed)`

`⇒ v = sqrt((2eV)/m_e`

The electron will move in a circular path due to the given magnetic field. Radius of the circular path,

`r = (m_ev)/(eB)`

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

`⇒ d > (m_e)/(eB)xx sqrt(2eV)/(m_e)`

`⇒ d > sqrt(2m_eV)/(eB^2)`

Thus, `d > ((2m_eV)/(eB_0^2))^{1/2}`