It has been experimentally observed that the electric field in a large region of the earth.s atmosphere is directed vertically down. At an altitude of 300m, the electric field is `60Vm^(-1)`. At an altitude of 200m, the field is `100Vm^(-1)`. Calculate the net amount of charge contained in the cube of 100m edge located between 200m and 300m altitude.

Explanation

Let `E_(1)` be electric field at 200m
and `E_(1)= 100V//m` (given), `E_(2)` be electric field at 300m and `E_(2)=60V//m`(given)
As wire of 100m edge is situated in electric field with its area normal to electric field, so flux through the cube at height of 200m is
`phi_(1)= E_(1).A= E_(1) A cos 0^(@)`
`=E_(1)A = 100 xx (100 xx 100)`
`=10 xx 10^(5) V-m`
Flux through the cube at height of 300m is
`phi_(2)= E_(2).A= E_(2)A cos 180^(@)`
`=-[60 xx (100 xx 100)]`
`= -6 xx 10^(5) V-m`
`therefore` Net flux through the cube is
`phi= phi_(1)+ phi_(2)`
`rArr phi= 10 xx 10^(5) -6 xx 10^(5) rArr phi= 4 xx 10^(5) V-m`
By Gauss.s theorem, `phi = (q)/(epsi_(0))` (where, q is net amount of charge)
`q= phi epsi_(0) = (4 xx 10^(5)) (8.85 xx 10^(-12))`
`=3.54 xx 10^(-6)C`