In given region electric and magnetic field are prpendicular to each other.Electron moving in this region do not get deflected,charge on cathode will be…..
[V=potential between anode and cathode]

Options

A. `(B^(2))/(2VE^(2))`
B. `(2VB^(2))/(E^(2))`
C. `(2VE^(2))/(B^(2))`
D. `(E^(2))/(2VB^(2))`

Answer

D

Explanation

f beam do not get deflected then eE=evB
`impliesv(E )/(B)` also `(1)/(2)mv^(2)=eV`
`therefore (e )/(m)=(v^(2))/(2V)=(E^(2))/(2VB^(2))`