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.

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

m_e = mass of the electron

From 3^rd 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