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

Options

A. `1.5xx10^(5)epsilon_(0)C`
B. `2xx10^(5)epsilon_(0)C`
C. `3xx10^(5)epsilon_(0)C`
D. `4xx10^(5)epsilon_(0)C`

Answer

d

Explanation

According to Gauss' theorem, electric flux:
`phi_(E)=(q)/(epsilon_(0))= int vec(E ).dvec(S)`
The surface integral in the above equations contains six terms- the surface intergal over the bottom surface, the surface intergal over the top surface and surface integral over the four vertical faces.
For the bottom surface, both the vectors `vec(E )` and `d vec(S)` are in the same direction. For the top surface, they act in opposite directions while for the vertical faces, they are perpendicular to each other. Hence,`phi_(E )=int_(bo t t om)vec(E )_(1).dvec(S)+int_(t op)vec(E )_(2).dvec(S)+4 int_(faces)vec(E ).dvec(S)=int_("bottom")E_(1).dS cos 0^(@)+int_(t op)E_(2).dS cos 180^(@)+4 int_(faces)EdS cos 90^(@)`
`=int_("bottom")E_(1).dS-int_("top")E_(2).dS=E_(1)S-E_(2)S=(E_(1)-E_(2))S`
Also, `phi_(E )=(q_(enclosed))/(epsilon_(0))`
`q=epsilon_(0)(E_(1)-E_(2))S= epsilon_(0)(100Vm^(-1)-60Vm^(-1))(100m)^(2)=4xx10^(5)epsilon_(0)C`