An electron is emitted with negligible speed from th nagative plate of a parallel plate capacitor charged ot a potential difference V. The separtion between the plates is d and a magnetic field B exists in the space as shown in . Show that the electron will fail to strike the upper plate if ltdgt `((2m_e V)/(eB_0^2))^(1/2)`.

Explanation

The force experienced first is due to the electric field due to the capacitor
`Since F. d = eV rArr q = (F)/(m) = (ev)/(m_(e) d)`
[where `e = charge of electron m_(e) = mass of electron]`
`v^(2) = u^(2) + 2` as
`rArr v_(2) = 2 xx ((eV)/(m_(e)) xx d`
`or v = sqrt((eV)/(m_(e))`
Force due to magnetic field product
`V = (eB_(0)d)/(m_(e))`
Now the electron will field strike upper plate only when `V` is greater then the `v`
Then `(eB_(0)d)/(m_(e)) gtsqrt((2m_(e)V)/(eB_(0) ^(2))`