The distance between the two plates of a cathode-ray oscilloscope is 1 cm and the potential difference between them is 1200 V. If an electron of energy 2000 eV enters at right angles to the field, what will be its deflection if the plate be 1.5 cm long.

Explanation

Electric field in between the plates is `E= (V)/(d) = ( 1200)/( 1 xx 10^(-2)) = 1.2 xx 10^(5) V//m `
Deflection can be obtained directly from equation of trajectory.
`Y _(0)(x=L) = ((qE)/( 2 m u^(2)))L^(2) = ( qE)/( 4K)L^(2)`
`implies ` Deflection `= ( qEL^(2))/( 4K)`
where `K =` Kinetic energy of electron
Given `: L = 1.5 xx 10^(-2) m`
`K = 2000 xx 1.6 xx 10^(-19) J `
`implies ` Deflecion `= 3.375 mm`