In the figure shown there is a large sheet of charge of uniform surface charge density`'sigma'`.A charge particle of charge `'-q'` and mass `'m'` is projected from a point A on the sheet with a speed `'u'` with the angle of projection such that it lands at maximum distance from A on the sheet. Neglecting gravity, find the time flight

Options

A. `(2sqrt(2)mvepsilon_(0))/(sigma q)`
B. `(sqrt(2)mv epsilon_(0))/(5 sigmaq)`
C. `(mv epsilon_(0))/(sqrt(2)sigmaq)`
D. `(mv epsilon_(0))/(2sqrt(2)sigma q)`

Answer

a

Explanation

For maximum distance angle of projection should be `45^(@)`

For vertical displacement, `vec(S)= vec(ut)+1/2vec(a)t^(2)`
`0= u sin 45^(@) t+1/2.(q.(sigma)/(2epsilon_(0)))/(m)t^(2)`
`t=0` or `t= (2sqrt(2)mvepsilon_(0))/(sigmaq)`
So time of flight `= (2sqrt(2)mvepsilon_(0))/(sigmaq)`