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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)`
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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 = 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}`
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