Question

An electron is fired directly towards the centre of a large metal plate that has excess negative charge with surface charge density 2.0 ✕ 10−6 C/m2. If the initial kinetic energy of the electron is 100 eV and if it is to stop (due to repulsion) just as it reaches the plate, how far from the plate must it be fired?

Answer 1

Inorder to stop an electron of 100eV energy to stop reaching charged metallic plate, the kinetic energy of electron must be equal to potential energy on the surface of metal plate.According to the Gauss's theorem electric field on the surface of a plate .E=σ2ε0.  Where σ is surface charge density and ε0 the permittivity of the free space.σ=2×10−6surface charge density/m2 and ε0 =8.85×10−12C2m−2N−1The potential energy at surface of the plate by the surplus negative charges is given by , U=e E∫r00dr       or U=[r]r00        or      U=e E r0.Therefore   U=  e×σ2ε0×r0.Nowkinetic energy of electron=100eV,   where  e electronic charge, r0 minimum distance of approach.Surface potential energy of charged plate =kinetic energy of approaching electron.So      e× σ2ε0×r0 =100eV,          or     r0  =2ε0×100σ                or                   r0 =2×8.85×10−12×1002×10−6= 8.85×10−4m

Answer 2

Answer:

.44mm

Explanation

simple you can put the formula [

Kinetic energy=Force ×displacement]

in question you have to find (d) .So

100×e- = e-×E×d

now ,d= [100/2×10^-6]×8.85×10^-12

by solving you will get your answer.