 # 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 the figure.Show that the electron will fail to strike the upper plate if d  > (2mV/eB2)1/2  ROSHAN MUJEEB
2 years ago
Potential difference across the plates of the capacitor =V
Separation between the plates =d
Magnetic field intensity =B
The electric field set up between the plates of a capacitor,E=VdE=Vd
The force experienced by the electron due to this electric field,F=eVdF=eVd
⇒a=Fm=eVmed⇒a=Fm=eVmed,
wheree= charge of the electron andme= mass of the electron
Usingv2=u2+ 2asand substituting the value ofa, we get:
v2=2×(eVmed)×d⇒v=2eVme−−−√v2=2×eVmed×d⇒v=2eVme
The electron will move in a circular path due to the given magnetic field. Radius of the circular path,
r=meveBr=meveB
And the electron will fail to strike the upper plate only when the radius of the circular path will be less thand,
i.e.d>r⇒d>meeB×2eVme−−−√⇒d>2meVeB2−−−−√i.e.d>r⇒d>meeB×2eVme⇒d>2meVeB2
Thus,d>(2meVeB02)12