There is uniformly charged disc of surface charge density `sigma` and of radius `1m`. `H=10m` height below the disc is a gun which is continuously emitting a charged particle and its isotopic in the vertically upward direction. The particles are charged through same potential before being emitted. The charge of the particle is `q=8xx10^(-19)C`. The difference of potential energies of particle and the isotopic at `B` is `10^(-23)`Joule. Also the velocity of the particle at `B` is zero. The ratio of masses of particle and isotopic is `(3)/(2)`. If the total energy of isotope at `B` is `1.1kxx10^(-23)J` Then find `k`..(Given `sigma//epsilon_(0)=10^(-4)N//C`).

Explanation

At `'A'` Total energy of particle
`E_(PA)=(1)/(2)m_(P)V_(PA)^(2)+qvarphi+0`
And of isotope
`E_(iA)=(1)/(2)m_(1)V_(PA)^(2)+qvarphi+0rArrE_(PA)=E_(iA)`
At `B`, `phi_(B)=(sigmaa)/(2epsilon_(0))`
Energy of particle is
`E_(PB)=(1)/(2)m_(P)xx0+qverphi_(a)+m_(P)gh`
Energy of istope `E_(iB)=(1)/(2)m_(i)V_(iB)^(2)+qvarphi_(B)+m_(i)gh`
`(m_(r)-m_(i))gh=10^(-23)`
`m_(P)=1.5xx10^(-25)kg`.
`m_(r)=0.5xx10^(-25)kg`.
`:. E_(PB)=E_(iB)`
`(1)/(2)m_(i)V_(B)^(2)=(m_(P)-m_(i))gh=10^(-23)`
`E_(iB)=10^(-23)+8xx0^(-19)xx0.5xx10^(-4)+(10^(-25))/(2)xx10xx10`
`=5.5xx10^(-23)J`.