Q. 52

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 d and a magnetic field B exists in the space as shown in figure. Show that the electron will fail to strike the upper plate if.



Answer :


Given-
Potential difference applied across the plates of the capacitor = V
Separation between the plates = d
Magnetic field intensity = B


The electric field applied across the plates of a capacitor


,



Also, coulomb’s force experienced by the electron is given by,



where e= charge on the electron and


E= electric field applied


Hence, the force experienced by the electron due to this electric field,



Now, acceleration a is given by-




where


e = charge of the electron


me = mass of the electron



From 3rd equation for motion



where


u = initial velocity


v= final velocity


s=distance travelled


and a = acceleration of the particle


substituting the value of a-




The electron will move in a circular path due to the presence of the magnetic field.


The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C


Radius of the circular path followed by the electron is ,



And the electron will fail to strike the upper plate of the


capacitor if and only if the radius of the circular path will be less


than d,


i.e. d>r




Thus, the electron will fail to strike the upper plate if



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