An electron enters the space between the plates of a charged capacity as shown. The charge density on the plate is `sigma`. Electric intensity in the space between the plates is E.A uniform magnetic field B also exists in that space perpendicular to the direction of E. The electron moves perpendicular to both `vec(E)` and `vec(B)` without any change in direction
The time taken by the electron to travel a distance l in that space is

Options

A. `(sigma l)/(epsilon_(0)B)`
B. `(sigmaB)/(epsilon_(0)l)`
C. `(epsilon_(0)lB)/(sigma)`
D. `(epsilon_(0)l)/(sigmaB)`

Answer

C

Explanation

Force on the electron due to the electric field E is
`F_(E)=(-e)E`
force on the electron due to the magnetic field B is
`F_(B)=(-e)vB`
The electron will move in the fields ubdeflected if these two forces are equal and opposite
`eE=evE` or `v=(E)/(B)`
Electric field between the plates is `E=(sigma)/(epsilon_(0))`
`:.v=(sigma)/(epsilon_(0)B)`
The time taken by the electron to travel a distance l in the space is
`t=(l)/(v)=(l)/((sigma)/(epsilon_(0)B))=(l epsilon_(0)B)/(sigma)`